Partial Differential Equations and Sobolev Spaces

MAT-INF4300 autumn 2014

Snorre H. Christiansen

November 12, 2014

1

Contents

1 Introduction 4

2 The heat equation 6

3 Exercises on partial derivatives 7

4 Continuous linear or bilinear maps 9

5 Laplace equation 10

6 Harmonic functions 11

7 Energy methods 127.1 Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127.2 The Dirichlet principle . . . . . . . . . . . . . . . . . . . . . . . . 127.3 Inner products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

8 Derivatives and completeness 158.1 Derivation and antiderivation . . . . . . . . . . . . . . . . . . . . 158.2 Completions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168.3 Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

9 Integration and mollification 199.1 Measures and integration . . . . . . . . . . . . . . . . . . . . . . 199.2 The standard bump . . . . . . . . . . . . . . . . . . . . . . . . . 219.3 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

10 Extension to L1(U) 2310.1 Epsilon–neighborhoods and epsilon–interiors . . . . . . . . . . . . 2310.2 Two results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

11 Lp spaces 2411.1 ... and convolution . . . . . . . . . . . . . . . . . . . . . . . . . . 2411.2 Exercises on Holder’s inequality . . . . . . . . . . . . . . . . . . . 2411.3 Complement: integral operators . . . . . . . . . . . . . . . . . . . 25

12 Weak derivatives 27

13 Lp spaces continued 28

14 Sobolev spaces 29

15 Distributions 30

16 Distributions continued 31

17 Stokes’ theorem 32

18 Approximation 33

19 No lecture 34

2

20 No lecture 35

21 Correction of last year’s compulsory exercises 36

22 Traces 3922.1 The trace operator W1,p(U)→ Lp(∂U) . . . . . . . . . . . . . . . 3922.2 Characterizations of W1,p

0 (U) . . . . . . . . . . . . . . . . . . . . 39

23 Poincare inequality and Dirichlet principle 4023.1 Poincare . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4023.2 Lax-Milgram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

24 Second order elliptic PDEs 4224.1 Variational formulation . . . . . . . . . . . . . . . . . . . . . . . 4224.2 Galerkin method . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

25 The space W1,1 of an interval and extension 4325.1 The space W1,1 of an interval . . . . . . . . . . . . . . . . . . . . 4325.2 The Cantor - Lebesgue function . . . . . . . . . . . . . . . . . . . 4525.3 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

26 Gagliardo-Nirenberg-Sobolev inequality 46

27 Rellich compactness 47

A Compulsory exercises 48

B Various exercises 50

3

1 Introduction

18 August 2014

Practical information. Prerequisites:

• Basic real analysis, e.g. MAT2400 (say compact subsets of Rn, uniformconvergence of sequences of functions).

• Aquaintance with PDEs, e.g. MAT-INF3360 (some exposure to partialderivatives including Gauss-Green-Stokes theorem, examples from physics,chemistry, biology, social sciences, finance or whatever).

Strongly recommended (either before or in parallel):

• Basic measure theory and functional analysis: MAT4400.

Curriculum: Lawrence C. Evans Partial Differential Equations, Amer. Math.Soc., Graduate Studies in Mathematics. Selected topics in the following chap-ters:

• Chap. 2: Some explicit formulas, for the heat and Poisson equations.

• Chap. 5: Sobolev spaces, definition and some properties.

• Chap. 6: Applications of Sobolev spaces to linear elliptic PDEs.

Time permitting: the abstract theory of finite element methods.

Language. A comparison of formulations. Let I = [a, b] be a (closed andbounded) interval in R. Consider the following statements:

• Suppose f, g : I → R are two functions. Pick x ∈ I. Suppose f and g areboth differentiable at x. Then f+g is differentiable at x and (f+g)′(x) =f ′(x) + g′(x).

• Let F denote the set of (real valued) functions on I and D denote theset of differentiable functions. Then F has a natural structure of vectorspace, and D is then a subspace. Moreover differentiation defines a linearmap from D to F .

• Denote by Ck(I) the space of functions which are k times differentiable onI, the derivatives up to order k being continuous. It is a vector space, andthe following defines a norm:

‖u‖k = sup{|u(l)(x)| : l ≤ k, x ∈ I}. (1)

Here u(l) denotes the derivative of order l of u. Then derivation definesa bounded linear map from Ck(I) to Ck−1(I) (k ≥ 1), by construction.Moreover Ck(I) equipped with this norm, is complete.

For this course I am going to assume that the students think the first state-ment is a distant memory, that they are familiar with the second statement andthat they can work through the last. This is mostly about reformulating keyresults from a second calculus course in the language of functional analysis.

Remarks:

4

• If the interval I is not compact, one can restrict attention to functionsthat are bounded and have bounded derivatives. This guarantees that theabove norm is still well defined.

• Similar statements hold for functions of several variables, as seen in class.

This defines some function spaces. Sobolev spaces are other function spaces,that turn out to be more adapted to the study of partial differential equations,for instance because Hilbert space techniques can be applied to them.

Substance. For most PDEs there is no “explicit formula” for the solution.One then studies questions such as: does the PDE have a solution, is it unique,and what are its properties. Consider the following reasoning:

Let U be an open bounded subset of Rn. Let f : U → R be a function andg : ∂U → R be another one. Consider the equation:

−∆u = f on U, (2)

u = g on ∂U. (3)

To show that this equation has at most one solution one writes:Suppose that u solves the homogeneous equation:

−∆u = 0 on U, (4)

u = 0 on ∂U. (5)

Then, by integration by parts:∫|∇u|2 = −

∫u∆u = 0. (6)

Hence ∇u = 0, so u is constant, and using the boundary condition again, wededuce u = 0. Since the equation is linear, this proves uniqueness.

There are many steps in this reasoning, and the course will justify each ofthem, and explore in which generality such results hold. For instance we needto define integration on the boundary of a set and justify integration by parts.In fact we will carry through this reasoning even for functions that are notdifferentiable! More precisely we will generalize the notion of derivative, so thatit applies to relatively “rough” functions, including some discontinuous ones.

For a numerical method applied to some PDE, one wants to estimate thedistance from the computed (approximate) solution to the exact solution, withrespect to some norm. It turns out that this is easier in Sobolev space normsthan in those, more elementary, defined by (1).

Welcome!

5

2 The heat equation

20 August 2014

I gave a derivation of the heat equation from “physical principles” and gavesome vocabulary related to several interpretations:

• temperature in a body (heat equation).

• density of a chemical in a fluid (convection–diffusion equation).

• value of an option (Black and Scholes).

Explicit formulas for the transport equation and the heat equation in ideal-ized scenarios.

Reading assignment, in [1]:

• Introduction.

• Chapter 2: §2.1 p. 18–19. §2.3 p. 44 – 48 (but the proof of Theorem 1 p.47 is optional).

• Appendices: Notations p. 697 onwards (get used to some of them...). Thedefinitions in §D.1 p. 719 (three first) and §D.3.a p. 721 (two first).

6

3 Exercises on partial derivatives

25 August 2014

Let U be an open subset of Rn.

Exercise 3.1. Check that if u ∈ C1(U) (scalar field) and v ∈ C1(U)n (vectorfield):

div(uv) = (gradu) · v + udiv v. (7)

Exercise 3.2. Choose reals ai < bi for integer i such that 1 ≤ i ≤ n. SupposeU =]a1, b1[× · · ·×]an, bn[. You may start with n = 2. What is the boundary ofU? What is the unit outward pointing normal on ∂U? We denote it by ν.

Give a direct proof of Stokes theorem in this case, namely that if u ∈C1(Rn)n then : ∫

U

div u =

∫∂U

u(x) · ν(x)dx. (8)

Deduce (informally) the formula for a finite union of such boxes, possibly sharingfaces.

Exercise 3.3. For a function u ∈ C1(U), we define for any vector a = (a1, · · · , an) ∈Rn:

∂au(x) = limε→0

ε−1(u(x+ εa)− u(x)). (9)

Check:

∂au(x) =

n∑i=1

ai∂iu(x). (10)

Let (ei)1≤i≤n denote the canonical basis of Rn:

e1 = (1, 0, 0, . . . , 0), (11)

e2 = (0, 1, 0, . . . , 0), (12)

. . . (13)

en = (0, 0, 0, . . . , 1). (14)

Compared with standard notation we have:

∂iu = ∂eiu. (15)

Let (fi)1≤i≤n be another orthonormal basis of Rn.Prove that for u ∈ C1(U):∑

i

(∂iu)ei =∑i

(∂fiu)fi. (16)

and that for u ∈ C1(U)n: ∑i

∂iui =∑i

∂fi(u · fi). (17)

Prove also that for u ∈ C2(U):

n∑i=1

∂2i u =

n∑i=1

∂2fiu. (18)

7

Exercise 3.4. Let A be an invertible real n×n matrix, considered also as a mapRn → Rn, x 7→ Ax. Check that for u ∈ C1(U) we have:

grad(u ◦A) = (AT gradu) ◦A, (19)

and that for v ∈ C1(U)n:

div(A−1v ◦A) = (div v) ◦A. (20)

Deduce that if A is an orthogonal matrix:

∆(u ◦A) = (∆u) ◦A. (21)

Whether A is orthogonal or not, find a positive definite matrix B such that:

∆(u ◦A) = (divB gradu) ◦A. (22)

8

4 Continuous linear or bilinear maps

28 august 2014

Exercise 4.1. [1] problem 13 p. 87.

Not very much functional analysis is required for this course. Essentiallywe need the following background material, as well as some results related tocompleteness that will be introduced later on.

In the following all vectorspaces are vectorspaces over a field K which iseither R or C.

Proposition 4.1. E and F two normed vectorspaces. A : E → F a linear map.The following are equivalent:

• A is continuous.

• A is continuous at 0.

• There is C > 0 such that for all u ∈ E, ‖Au‖ ≤ C‖u‖.

Such linear maps are also said to be bounded.

Proof. See Lemma 4.1 p. 88 in [7].

For a continuous linear map A : E → F one defines:

‖A‖ = inf{C ≥ 0 : ∀u ∈ E ‖Au‖ ≤ C‖u‖}, (23)

= sup{‖Au‖ : u ∈ E ‖u‖ ≤ 1}. (24)

The continuous linear maps E → F form a vector space denoted L(E,F )and the above expression defines a norm on this space.

In the special case F = K, the space L(E,K) is called the dual space of Eand denoted E?.

Very similarly one proves:

Proposition 4.2. E, F and G three normed vectorspaces. a : E × F → G abilinear map. The following are equivalent:

• a is continuous.

• a is continuous at 0.

• There is C > 0 such that for all u ∈ E, all v ∈ F , ‖a(u, v)‖ ≤ C‖u‖ ‖v‖.

Such bilinear maps are also said to be bounded.For a continuous bilinear map a : E × F → G one defines:

‖a‖ = inf{C ≥ 0 : ∀u ∈ E ∀v ∈ F ‖a(u, v)‖ ≤ C‖u‖ ‖v‖}, (25)

= sup{‖a(u, v)‖ : u ∈ E ‖u‖ ≤ 1, v ∈ F ‖v‖ ≤ 1}. (26)

The continuous bilinear maps constitute a normed vector space.Notice that a continous linear map is uniformly continuous, whereas a (non–

zero) continuous bilinear map is not.

9

5 Laplace equation

1 september 2014

• Multi-index notation for partial derivatives.

• Differential operators.

• Support of a continuous function.

• Spherically symmetric solutions of the Laplace equation.

• Fundamental solution.

10

6 Harmonic functions

4 september 2014

• Convolution.

• Solving the Poisson equation using the fundamental solution.

• Mean value property.

• Standard mollifier.

• Smoothness of harmonic functions.

• Liouville theorem.

• An existence and uniqueness result for solutions to Poisson (n ≥ 3).

Reading assignment: Evans Section 2.2.1, 2.2.2, 2.2.3 (b ,d). See also Ap-pendix C.5, E.1, E.2, E.3.

11

7 Energy methods

8 september 2014

7.1 Complements

PDE theory draws on many (all?) sources. I hope you will find references thatplease you. Here are some of my favorites.

Differential Intrinsic definition of the differential of a function from onenormed vector space to another, compared with partial derivatives:

• [10] Chapter 17: Several variable differential calculus.

• [5] Chapter XV: Functions on n-space,and Chapter XVII: Derivatives in vector spaces.

Lebesgue measure and integral

• [9] Chapter 1: Measure theory.

• [2] Chapter 2.6: The n-dimensional Lebesgue integral.

Convolution Convolution and Dirac sequences:

• [5] Chapter XI: Approximation with convolution.

• [9] Chapter 3.2: Good kernels and approximations to the identity.See also: [8] Chapter 2.3: Convolutions, Chapter 2.4: Good kernels.

• [2] Chapter 8.2: Convolutions.

7.2 The Dirichlet principle

Already known to Gauss, but used and extended by Riemann under this name.Reading assignment: §2.2.5 in Evans.

7.3 Inner products

In the following all vector spaces are real.The following is the Cauchy-Schwarz inequality (we check that it doesn’t

require positive definiteness).

Proposition 7.1. Let X be a vector space and a a symmetric bilinear form onX with the property:

∀u ∈ X a(u, u) ≥ 0. (27)

Then:∀u, v ∈ X |a(u, v)| ≤ a(u, u)1/2a(v, v)1/2, (28)

and the function: {X → R,u 7→ a(u, u)1/2.

, (29)

defines a seminorm on X.

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Proof. (i) Pick u, v ∈ X, for which we want to prove (28).We first write for all λ ∈ R:

0 ≤ a(u+ λv, u+ λv) = a(u, u) + 2λa(u, v) + λ2a(v, v). (30)

With λ = −sgna(u, v) we get:

2|a(u, v)| ≤ a(u, u) + a(v, v). (31)

From this we deduce, for all t > 0:

2|a(u, v)| = 2|a(tu, t−1v)| ≤ t2a(u, u) + t−2a(v, v). (32)

If both a(u, u) = 0 and a(v, v) = 0, then a(u, v) = 0 and we are done.Suppose then, that a(u, u) 6= 0. If a(v, v) = 0 we get a(u, v) = 0 by letting

t→ 0, so we are also done. If a(v, v) 6= 0, choosing t so that:

t2 = a(v, v)1/2/a(u, u)1/2, (33)

gives the result.(ii) We have:

a(u+ v, u+ v)1/2 = (a(u, u) + 2a(u, v) + a(v, v))1/2, (34)

≤ (a(u, u) + 2a(u, u)1/2a(v, v)1/2 + a(v, v))1/2, (35)

≤ a(u, u)1/2 + a(v, v)1/2. (36)

This is the main ingredient in showing that (29) defines a seminorm.

In the above circumstaces, the seminorm is a norm iff:

a(u, u) = 0⇒ u = 0. (37)

In this case we call a an inner product on X and we say that (X, a) is aninner product space. When X is equipped with the norm defined by a, a is acontinuous bilinear form on X.

Definition 7.1. A Hilbert space is an inner product space which is complete.

Recall that for any normed vector space X, its dual is the space of continuouslinear forms on X. We denote the dual by X?. It is equipped with the norm:

‖l‖ = supu∈X

|l(u)|‖u‖

, (38)

where it is understood that only non-zero u are used.

Proposition 7.2. Let (X, a) be an innerproduct space and l ∈ X?. Let F :X → R denote the functional defined by:

F (u) =1

2a(u, u)− l(u). (39)

For any u ∈ X, the following are equivalent:

∀v ∈ X a(u, v) = l(v). (40)

∀v ∈ X F (u) ≤ F (v). (41)

There is at most one such u.

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Remark 7.1. Let (X, a) be an inner product space. Then, for u ∈ X:

‖u‖ = supu∈X

|a(u, v)|‖v‖

= ‖a(u, ·)‖X? . (42)

In other words the linear map:{X → X?,u 7→ a(u, ·). (43)

is an isometry.

The following result is called the Riesz representation theorem.

Theorem 7.3. Let (X, a) be a Hilbert space. Then for any l ∈ X? there existsa unique u ∈ X such that:

∀v ∈ X a(u, v) = l(v). (44)

Proof. Let F be the functional defined in (39).For any v ∈ X we have:

F (v) ≥ 1

2‖v‖2 − ‖l‖ ‖v‖, (45)

≥ 1

2(‖v‖ − ‖l‖)2 − 1

2‖l‖2 ≥ −1

2‖l‖2. (46)

Define:I = inf{F (v) : v ∈ X}. (47)

Let (un) be a sequence in X such that:

F (un)→ I. (48)

(one says that (un) is a “minimizing sequence”).We have (from the “parallelogram identity”):

1

4‖un − um‖2 =

1

2‖un‖2 +

1

2‖um‖2 − ‖

un + um2

‖2, (49)

= F (un)− l(un) + F (un)− l(un)− (50)

2(F (un + um

2)− l(un + um

2)), (51)

= F (un) + F (um)− 2F (un + um

2), (52)

≤ F (un)− I + F (um)− I. (53)

It follows that (un) is Cauchy (notice how the fact that I > −∞ enters theargument). Let u be its limit. By continuity of F , F (u) = I, so F achieves itsminimum at u.

Remark 7.2. Let (X, a) be an inner product space. It follows from the com-pleteness of R, that X? is complete1, so in order for the map (43) to be onto,it is necessary that (X, a) is complete. The theorem says that this condition isalso sufficient.

1More generally, if X is a normed vector space any Y is a Banach space, then the spaceL(X,Y ) of continuous linear maps X → Y , equipped with the operator norm, is a Banachspace.

14

8 Derivatives and completeness

10 september 2014

8.1 Derivation and antiderivation

The following can be called the Lemma of bounded increments.

Lemma 8.1. Let X be a normed vector space. Suppose that f : [a, b] → X iscontinuous on [a, b] and differentiable on ]a, b[, with:

∀x ∈]a, b[ ‖f ′(x)‖ ≤M. (54)

Then:‖f(b)− f(a)‖ ≤M |b− a|. (55)

Proof. The trick is to prove that for any ε > 0, the set of points x ∈ [a, b] suchthat:

‖f(x)− f(a)‖ ≤ ε+ (M + ε)|x− a|, (56)

contains b.

Corollary 8.2. Let X be a normed vector space. Suppose that f : [a, b] → Xis continuous on [a, b] and differentiable on ]a, b[, with differential 0. Then f isconstant.

Proposition 8.3. Let X be a complete normed vector space. Suppose I is aninterval and a ∈ I. Suppose fn : I → X is a sequence of differentiable functionssuch that:

• fn(a) converges.

• f ′n converges uniformly to a function g : I → X.

Then fn converges pointwise to a function f : I → X, and the convergence isuniform on any bounded part of I. Moreover f is differentiable, with derivativeg.

Proposition 8.4. Let I be an interval and X a complete normed vector space.Consider the space of differentiable functions I → X, which are bounded withbounded derivative. Equip it with the norm:

‖f‖ = supx∈I‖f(x)‖+ sup

x∈I‖f ′(x)‖. (57)

This space is complete.Moreover the continuously differentiable functions which are bounded with

bounded derivative, constitute a closed subspace.

Theorem 8.5. Let X be a complete normed vector space. Suppose I is aninterval and that f : I → X is continuous. Then there is a a function F : I → Xwhose derivative is f , and it is unique up to a constant.

Proof. (Sketch). Suppose the interval is closed and bounded and choose a ∈I. Approximate f uniformly by a sequence fn of continuous piecewise linearfunctions. Let Fn be an antiderivative of fn which has value 0 at a (it existsbecause we have an explicit formula). Then apply Proposition 8.3.

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Remark 8.1. This gives a first possible approach to integration. For a continuousfunction f : [a, b]→ X, choose an antiderivative F and define:∫ b

a

f = F (b)− F (a). (58)

This can be extended to piecewise continuous functions.

Exercise 8.1. Let I be a bounded closed interval. We have the space C1(I)equipped with the standard norm. Another norm on this space is the onederived from the scalar product:

(u, v)→∫uv +

∫u′v′. (59)

Find a sequence in C1(I) which is Cauchy with respect to the latter norm, butnot the former.

The following is Weierstrass critique of the Dirichlet principle.

Exercise 8.2. Define:

A = {v ∈ C1([−1, 1]) : v(−1) = −1, v(1) = 1}. (60)

We define a functional J : A → R by:

J(v) =

∫ 1

−1x2|v′(x)|2dx. (61)

a) Prove that:

∀v ∈ A J(v) > 0. (62)

b) Define a function vn by:

vn(x) =arctan(nx)

arctan(n). (63)

Show that J(vn)→ 0.c) Deduce that there is no point in A where J reaches its infimum.

8.2 Completions

Lemma 8.6. E and F two normed vectorspaces. Suppose E0 a linear subspaceof E which is dense. Suppose A0 : E0 → F a linear map. Suppose that F iscomplete and that there is C > 0 such that:

∀u ∈ E0 ‖A0u‖ ≤ C‖u‖. (64)

Then there is a unique continuous map A : E → F whose restriction to E0 isA0. It is linear and satisfies:

∀u ∈ E ‖Au‖ ≤ C‖u‖. (65)

Proof. [5] Chapter X.1, Theorem 1.2 p. 247.

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Definition 8.1. Suppose X is normed vector space. A completion of X is acomplete normed vectorspace Y together with a linear isometry i : X → Ywhose range is a dense subspace of Y .

Example 8.1. If X is a dense subspace of some complete space Y , the norm ofX being the restriction of the norm of Y , take i : X → Y to be the canonicalinjection.

The following is a uniqueness result for completions.

Lemma 8.7. If, given X, we have two completions (i, Y ) and (j, Z) there is aunique continuous map h : Y → Z such that h ◦ i = j. Moreover h is a linearbijective isometry.

Theorem 8.8. Any normed vector space has a completion.

Proof. (Sketch) Suppose we want to complete a normed vector space X. Onetakes the space of Cauchy sequences in X, equipped with the sup norm. This is aBanach space. Those sequences that converge to 0 constitute a closed subspace.The quotient space is our space Y and we take for i : X → Y the map sendingan element u in X to the equivalence class of the constant sequence with valuesu.

See [5] Chapter VII.4: Completion of a normed vector space. See also [10]Exercise 12.4.8.

But this particular “canonical” completion of a space, if it is applied to aspace of functions, does not yield a space of functions. In this case one wouldlike to know if the completion can be identified with a space of functions or not.

8.3 Complements

Integration of regulated functions The following sketches a way of defin-ing the integral based on completion. For details on this approach, see [5]Chapter X: The integral in one variable.

Let [a, b] be an interval and X a complete normed vector space. We take forgranted the following notions:

• Step functions from [a, b] to X.

• Integration of step functions (independence of subdivision).

Definition 8.2. A function f : [a, b] → X is said to be regulated if it is theuniform limit of step functions.

Proposition 8.9. The space of regulated functions, equipped with the sup-normis complete. The step functions constitute a dense subspace.

Proposition 8.10. There is a unique continuous linear form on the space ofregulated functions, whose restriction to the step functions is the integral.

Proposition 8.11. Any continuous function is regulated.

Theorem 8.12. Let f : [a, b]→ X be a continuous function.

17

• Define F : [a, b]→ X by:

F (x) =

∫ x

a

f(y)dy. (66)

Then F is differentiable and F ′ = f .

• Suppose F is any antiderivative of f . Then:∫ b

a

f = F (b)− F (a). (67)

Remark 8.2. For those who know the Riemann integral: Any regulated functionis Riemann integrable (easy), but the converse is false (hard).

Remark 8.3. Consider a regulated function f . Can we extend Theorem 8.12 tosuch functions? For a complete answer, see [3] Chapter V.3.13: Extension dutheoreme fondamental aux fonctions reglees.

An approach to the Lebesgue integral through completion Let S de-note the space of step functions [a, b]→ X. Equip S with the seminorm definedby:

|f | =∫ b

a

‖f(x)‖dx. (68)

It turns out that the completion of this space with respect to this semi-norm,has all the properties we would like for a nice space L1([a, b], X). In particularits elements can be identified with functions (defined up to a subset of measurezero).

Realizing this program requires some work of course, see [5] Chapter X.4p. 262 – 267, and [6] Appendix A, p. 529 – 541. Remark that completing Swith respect to the sup-norm yields the space of regulated functions and wascomparatively easy.

18

9 Integration and mollification

14 september 2014

9.1 Measures and integration

Definition 9.1. Let E be a set. A σ-algebra on E is a subsetM of P(E) suchthat:

• ∅ ∈ M,

• For all F ∈M we have E \ F ∈M,

• For any sequence (Fn)n∈N in M we have ∪n∈NFn ∈M.

When E is a topological space, there is a smallest σ-algebra containing theopen subsets of E, called the Borel σ-algebra on E.

Definition 9.2. Let E be a set and M a σ-algebra on E. A measure on M isa function µ :M→ [0,∞] such that for any sequence (Fn)n∈N in M of two bytwo disjoint sets:

µ(∪n∈NFn) =∑n∈N

µ(Fn). (69)

Let E be a set, M a σ-algebra on E and µ a measure on M. Then thefollowing is also a σ-algebra:

M = {F ∪G : F ∈M, and ∃G′ ∈M µ(G′) = 0 and G ⊆ G′}. (70)

The measure µ has a unique extension to a measure µ on M . For any subsetF of E, if there is F ′ ∈ M such that µ(F ′) = 0 and F ⊆ F ′, we have F ∈ M.One says that (M,µ) is the completion of (M,µ).

Once a σ-algebra on E has been fixed, as well as a measure, the elements ofthe σ-algebra will be referred to as the measurable subsets of E.

A property P(x) defined for x ∈ E is said to hold almost everywhere, if thereis a set F of measure 0 such that P(x) is true for x ∈ E \ F .

Definition 9.3. A function f : E → R is measurable if for any open interval Iin R, f−1(I) is a measurable subset of E.

Definition 9.4. For a function f : E → R the following are equivalent:

• f is measurable, has finite range and for a 6= 0, f−1({a}) has finite mea-sure.

• f is a (finite) linear combination of characteristic functions of measurablesubsets with finite measure.

Such functions are called simple. They form a vector space.

Lemma 9.1. Suppose f is a simple function represented as:

f =

n∑i=1

aiχFi , (71)

19

with Fi measurable, with finite measure. If f(x) = 0 for almost every x ∈ E,then:

n∑i=1

aiµ(Fi) = 0. (72)

As a consequence, if a simple function f is represented as in (71), the number:

n∑i=1

aiµ(Fi), (73)

does not depend on the choice of representation. It is called the integral of fand denoted

∫f .

For a measurable function f : E → [0,∞] we define:∫f = sup{

∫φ : φ is simple and φ ≤ f} ∈ [0,∞]. (74)

We define L1(E) to consist of measurable functions f : E → R such that‖f‖ =

∫|f | <∞. It is a normed vector space and the simple functions are dense

in it. The integral, defined a priori for simple functions, has a unique extensionto a continuous linear form on L1(E) called the integral and still denoted

∫.

Theorem 9.2. (dominated convergence) Suppose that f : E → R is somefunction, (fn)n∈N is a sequence in L1(E) and g : E → [0,∞] satisfies:

• g is measurable and∫g <∞,

• for almost every x ∈ E, for all n ∈ N, |fn(x)| ≤ g(x) ,

• for almost every x ∈ E, fn(x)→ f(x) as n→∞.

Then f ∈ L1(E) and:

limn→∞

∫fn =

∫f. (75)

Proposition 9.3. L1(E) is complete.

Actually the norm on L1(E) is just a seminorm, since the L1(E) norm of afunction is 0 iff the function is zero almost everywhere. One usually overlooksthis problem and considers that such functions are 0.

Proposition 9.4. On the Borel σ-algebra of Rn there is a unique measurewhich:

• is translation invariant,

• and is such that the measure of the open unit cube is 1.

The completion of this measure is called the Lebesgue measure and theassociated integral is called the Lebesgue integral. By default this is the measureand integral we will always consider.

If S is an open subset of Rn we denote by L1(S) the space of functions on Swhose extension by 0 to Rn is in L1(Rn). This is also a Banach space.

Proposition 9.5. For any open subset S of Rn, the space Cc(S) is dense inL1(S).

20

We define:

L1(S)loc = {u : Rn → R : ∀K ⊂ S K compact ⇒ χKu ∈ L1(S)}. (76)

For instance the fundamental solution to the Laplace operator is in L1(Rn)locbut not L1(Rn).

9.2 The standard bump

Define a function φ : R→ R by, for x > 0 :

φ(x) = exp(−1/x), (77)

and for x ≤ 0, φ(x) = 0.

Exercise 9.1. By induction on k, show that for any k ∈ N there exists a poly-nomial P and l ∈ N such that the k-th order derivative of φ takes the form, forx > 0:

φ(k)(x) =P (x)

xlexp(−1/x). (78)

Deduce that φ ∈ C∞(R).

Define Φ : Rn → R by, for x ∈ Rn:

Φ(x) = φ(1− |x|2). (79)

What is the support of Φ? Why is its integral strictly positive?Define then:

µ(x) = Φ(x)/∫

Φ, (80)

and:

µε(x) = ε−nµ(ε−1x). (81)

What is the integral of µε? These functions will be called the standardmollifiers.

9.3 Convolution

Given two functions u, v : Rn → R one defines u ∗ v : Rn → R by:

u ∗ v(x) =

∫u(y)v(x− y)dy, (82)

whenever this makes sense.

Proposition 9.6. For u, v ∈ L1(Rn) we get a well defined u ∗ v ∈ L1(Rn) and:

‖u ∗ v‖L1(Rn) ≤ ‖u‖L1(Rn)‖v‖L1(Rn). (83)

Exercise 9.2. Check that u ∗ v = v ∗ u.

Lemma 9.7. If u ∈ Cc(Rn), then u is uniformly continuous.

Proposition 9.8. If u ∈ L1loc(Rn) and v ∈ Cc(Rn) then u ∗ v ∈ C(Rn).

21

Proposition 9.9. If u ∈ L1loc(Rn) and v ∈ C1c (Rn) then u ∗ v ∈ C1(Rn) and:

∂i(u ∗ v) = u ∗ ∂iv. (84)

Proposition 9.10. If u ∈ Cc(Rn) then as ε→ 0, u∗Ψε converges to u, uniformlyon Rn.

Proposition 9.11. If u ∈ L1(Rn) then as ε → 0, u ∗ Ψε converges to u inL1(Rn).

Corollary 9.12. If u ∈ L1loc(Rn) satisfies, for all v ∈ C∞c (Rn):∫

uv = 0, (85)

then u = 0.

22

10 Extension to L1(U)

17 september 2014

• Correction of the exercises.

10.1 Epsilon–neighborhoods and epsilon–interiors

For a subset E of Rn and a point x ∈ Rn one defines:

d(x,E) = inf{|x− y| : y ∈ E}. (86)

Exercise 10.1. Check:d(x,E) = 0 ⇐⇒ x ∈ E, (87)

and:|d(x,E)− d(y,E)| ≤ |x− y|. (88)

Definition 10.1. Let E be a subset of Rn:

• The ε–neighborhood of E is:

Nε(E) = {x ∈ Rn : d(x,E) < ε}. (89)

• The ε–interior of a set E is:

Iε(E) = {x ∈ E : d(x,Rn \ E) > ε}. (90)

Lemma 10.1. If ε < ε′ then:

N ε(E) ⊆ Nε′(E). (91)

Lemma 10.2. If K ⊆ U ⊆ Rn, with K compact and U open, there exists ε > 0such that Nε(K) ⊆ U .

Exercise 10.2. supp(v ∗ µε) ⊆ Nε(supp(v)).

Exercise 10.3. Nε(Iε(E)) ⊆ E.

10.2 Two results

Proposition 10.3. Suppose U is open in Rn and u ∈ L1loc(U) satisfies:

∀v ∈ C∞c (U)

∫uv = 0. (92)

Then u = 0.

Proposition 10.4. Suppose U is open in Rn . Then C∞c (U) is dense in L1(U).

23

11 Lp spaces

22 september 2014

11.1 ... and convolution

• A reference for the curious: [2] Section 6.1 (p. 181 – 186): Basic theory of Lp

spaces.

Proposition 11.1. For functions on Rn:

‖u ∗ v‖Lp ≤ ‖u‖L1‖v‖Lp . (93)

A useful variant is:

Proposition 11.2. For a function defined on U and the standard mollifier µε:

‖u ∗ µε‖Lp(Iε(U)) ≤ ‖u‖Lp(U). (94)

Proposition 11.3. Suppose U is open in Rn. Then C∞c (U) is dense in Lp(U)for p ∈ [1,∞[.

11.2 Exercises on Holder’s inequality

Exercise 11.1. Suppose U ⊆ Rn is open. Suppose we have reals p, q ≥ 1 suchthat:

1

p+

1

q= 1. (95)

Given u ∈ Lp(U), find a non-zero v ∈ Lq(U) such that:∫uv = ‖u‖Lp(U)‖v‖Lq(U). (96)

Hint: find v such that uv = |u|p.

Exercise 11.2. Suppose U ⊆ Rn is open and has finite measure (e.g. bounded).Suppose we have reals 1 ≤ p < q. Suppose u ∈ Lq(U). Show that u ∈ Lp(U)and:

‖u‖Lp(U) ≤ (measU)(1p−

1q )‖u‖Lq(U). (97)

Hint: write |u|p = 1 · |u|p and find a Holder inequality that can be appliedto this product.

Exercise 11.3. Suppose U ⊆ Rn is open and that reals p, q, r ≥ 1 satisfy:

1

p+

1

q=

1

r. (98)

Suppose u ∈ Lp(U) and v ∈ Lq(U). Show that uv ∈ Lr(U) and:

‖uv‖Lr(U) ≤ ‖u‖Lp(U)‖v‖Lq(U). (99)

24

Exercise 11.4. Suppose U ⊆ Rn is open and that reals p, q, r ≥ 1 satisfy:

1

p+

1

q+

1

r= 1. (100)

Suppose u ∈ Lp(U), v ∈ Lq(U) and w ∈ Lr(U). Show that uvw ∈ L1(U) and:

‖uvw‖L1(U) ≤ ‖u‖Lp(U)‖v‖Lq(U)‖w‖Lr(U). (101)

Exercise 11.5. Suppose U ⊆ Rn is open and that we have reals p, q ≥ 1, as wellas θ ∈ [0, 1]. Define r by:

1

r= θ

1

p+ (1− θ)1

q. (102)

Suppose u ∈ Lp(U) ∩ Lq(U). Show that u ∈ Lr(U) and:

‖u‖Lr(U) ≤ ‖u‖θLp(U)‖u‖(1−θ)Lq(U). (103)

Definition 11.1. Suppose U ⊆ Rn is open. For u : U → R measurable, onedefines

‖u‖L∞(U) = inf{C ∈ [0,∞] : for a.e. x ∈ U |u(x)| ≤ C}. (104)

One also defines:

L∞(U) = {u : U → R : u is measurable and ‖u‖L∞(U) <∞}. (105)

One checks that this is a vectorspace and that (104) defines a norm on thisspace. Moreover this normed vectorspace is complete.

Exercise 11.6. Try to extend the preceding exercises to the case where at leastone of p, q and r is infinite.

11.3 Complement: integral operators

In the following (as always) we only consider integration with respect to Lebesguemeasure.

Theorem 11.4. Suppose U ⊆ Rn and V ⊆ Rm are open. Suppose K : U×V →R is measurable, and that we have two constants 0 ≤ C0, C1 <∞ such that:

for a.e. x ∈ U∫|K(x, y)|dy ≤ C0. (106)

and

for a.e. y ∈ V∫|K(x, y)|dx ≤ C1. (107)

Pick p ∈ [1,∞] and define q ∈ [1,∞] by:

1

p+

1

q= 1. (108)

Pick u ∈ Lp(V ). Then the following holds.

25

For a.e. x ∈ U the function y 7→ K(x, y)u(y) is in L1(V ). Moreover theexpression:

(Ku)(x) =

∫V

K(x, y)u(y)dy, (109)

defines a function Ku ∈ Lp(U) which satifies:

‖Ku‖Lp(U) ≤ C1q

0 C1p

1 ‖u‖Lp(V ). (110)

If one does not worry about which functions are measurable, this is just aclever application of Holder’s inequality. The measure-theoretic details of theproof, which involves an appeal to the Fubini and Tonelli theorems, are beyondthe scope of this course. Reference: [2] Theorem 6.18.

26

12 Weak derivatives

24 september 2014

Weak derivatives. See [1] §5.2.

27

13 Lp spaces continued

29 september 2014

• Correction of the exercises on Holder.

28

14 Sobolev spaces

1 october 2014

• Finished [1] §5.2.

29

15 Distributions

6 october 2014

Introduction to distributions (in the sense of L. Schwartz).A reference: [2] §9.1.The canonical reference is now probably [4], but a bit too long for a course...

• Definition of distribution. Order.

• Examples: distribution associated with a L1loc function. Dirac deltas.

Principal value of x 7→ 1/x.

• Multiplication by smooth functions. Associativity.

• Derivation. A Leibniz rule for products.

• Derivation of functions with a jump discontinuity.

• Convergence in the sense of distributions. Example: the standard molli-fiers.

30

16 Distributions continued

8 october 2014

• Correction of exercise: PV (x 7→ 1/x) defines a distribution.

• Pathologies associated with attempts to multiply distributions.

• Partitions of unity.

• Restriction and support of a distribution.

31

17 Stokes’ theorem

13 october 2014

• Domains

• Integration on boundaries

• Piola transform

• Stokes theorem

32

18 Approximation

15 october 2014

Reference: [1] §5.3.Evans presents three theorems on the approximation of elements in Sobolev

spaces by smooth functions.

• Theorem 1 says that as long as one stays away from the boundary, smooth-ing by convolution will do the job.

• Theorem 2 shows how one can get global approximation by smooth func-tions defined on the whole of U .

• Theorem 3 shows how one can get global approximation by smooth funci-tons defined on a neighborhood of U . Notice that contrary to Theorem 2,this uses some smoothness assumption on ∂U .

Of these three, the most useful one is the last, because it shows that there aredense subspaces of W1,p(U) for which integration by parts makes sense. This isused to define traces of functions on the boundary (See Evans §5.5).

33

19 No lecture

20 october 2014

Work on the compulsory exercises (Appendix A).

34

20 No lecture

22 october 2014

Work on the compulsory exercises (Appendix A).

35

21 Correction of last year’s compulsory exer-cises

27 october 2014

The following exercises constituted a subset of the compulsory exercises forautumn 2013. They will be corrected this day.

Some notation: For x ∈ Rn:

|x| = (x21 + · · ·+ x2n)1/2. (111)

and for an open U ⊆ Rn and a measurable vectorfield v : U → Rn we define:

‖v‖Lp(U) = ‖ |v| ‖Lp(U). (112)

Exercise 21.1. We define:

R+ = {x ∈ R : x > 0}. (113)

For any continuous function w : Rn → R+ define the space:

Xw = {u : Rn → R : u is measurable and

∫u2w <∞}. (114)

For u ∈ Xw we define:

‖u‖w = (

∫u2w)1/2. (115)

(a) Fix a continuous function w : Rn → R+. Check that Xw is a vector spaceand that ‖ ·‖w defines a norm on this space. In the following we always considerXw as equipped with this norm. Show that for any open bounded subset U ofRn we have a continous linear surjection:

Xw → L2(U), (116)

defined by u 7→ u|U .(b) Pick two functions w0 and w1, which are continuous Rn → R+. For

θ ∈]0, 1[ define:wθ = w1−θ

0 wθ1. (117)

Show that for any θ ∈]0, 1[ and any u ∈ Xw0∩Xw1

we have u ∈ Xwθ and:

‖u‖wθ ≤ ‖u‖1−θw0‖u‖θw1

. (118)

(c) Consider the setup of question (b). Suppose we have, for each n ∈ N afunction un ∈ Xw0

∩Xw1. Suppose that the sequence (un) converges in Xw0

tou ∈ Xw0 . Suppose also that (un) is bounded in Xw1 . Check that there exists aconstant C ≥ 0 such that for any open bounded subset U of Rn:∫

U

u2w1 ≤ C. (119)

Use the monotone convergence theorem to deduce that u ∈ Xw1. Show that

(un) converges to u in Xwθ for any θ ∈]0, 1[.

36

Exercise 21.2. (a) Show that there exists a function θ ∈ C∞c (Rn) such that:

∀x ∈ Rn |x| ≤ 1⇒ θ(x) = 1. (120)

(b) Let θ be a function as in (a). For non-zero m ∈ N define θm : Rn → Rby :

θm(x) = θ(x/m). (121)

Fix p ∈ [1,+∞[.- Show that for any u ∈ Lp(Rn), the functions θmu are in Lp(Rn) and

converge to u in Lp(Rn), as m→∞.- Show that for any u ∈ W1,p(Rn), the functions θmu are in W1,p(Rn) and

converge to u in W1,p(Rn), as m→∞.(c) Use known results on smoothing by convolution to deduce that C∞c (Rn)

is dense in W1,p(Rn) for all p ∈ [1,+∞[.

Exercise 21.3. Pick a real c > 0. Given a function v : R3 → R we consider theequation, on R3:

−∆u+ c2u = v. (122)

Given v ∈ C(R3), a strong solution to the equation (122) is a functionu ∈ C2(R3) such that (122) holds in the sense of pointwise differentiation.

Given v ∈ L1loc(R3), a weak solution to the equation (122) is a function

u ∈ L1loc(R3) such that for all φ ∈ C∞c (R3):∫

u(−∆φ+ c2φ) =

∫vφ. (123)

(a) Show that if v ∈ C(R3), a function u ∈ C2(Rn) is a strong solutionto (122) if and only if it is a weak solution. Also show that the homogeneousequation (that is, with v = 0) always has infinitely many strong solutions.

(b) Find all functions f ∈ C2(]0,+∞[) such that the function uf : R3\{0} →R defined by,

∀x ∈ R3 \ {0} uf (x) = f(|x|), (124)

satisfies (in the sense of pointwise differentiation):

−∆uf + c2uf = 0 on R3 \ {0}. (125)

(c) Define the function E : R3 → R by:

E(x) =1

4π|x|exp(−c|x|). (126)

Check that E is locally integrable and show that for all φ ∈ C∞c (R3):∫E(−∆φ+ c2φ) = φ(0). (127)

Deduce that for any v ∈ C∞c (R3), E ∗ v is a strong solution to (122).(d) Show that E ∈ W1,1(R3). Deduce that for v ∈ L2(R3), E ∗ v ∈ H1(R3)

and is a weak solution to (122).

37

(e) For this question you may use the result of question (c) in Exercise 21.2.Pick v ∈ L2(R3) and u ∈ H1(R3). Show that u is a weak solution to (122) ifand only if for any w ∈ H1(R3):∫

gradu · gradw + c2uw =

∫vw. (128)

Deduce that, given v ∈ L2(R3), equation (122) has a unique weak solution inH1(R3).

Exercise 21.4. For any function u : Rn → R and x ∈ Rn we define τxu : Rn → Rby:

∀y ∈ Rn (τxu)(y) = u(y − x). (129)

Show that for any u ∈ H10(Rn) and any x ∈ Rn:

‖u− τxu‖L2(Rn) ≤ |x| ‖ gradu‖L2(Rn). (130)

You may start by noticing that for u ∈ C∞c (Rn):

u(y)− u(z) =

∫ 1

0

gradu(z + t(y − z)) · (y − z)dt. (131)

38

22 Traces

29 october 2014

Reference: [1] §5.5.

22.1 The trace operator W1,p(U)→ Lp(∂U)

Notice that there are essentially three ingredients in its definition:

• A lemma on the construction of linear operators (extension by continuityfrom dense subspaces): Lemma 8.6 in these notes.

• A so-called “estimate” on functions, proved for smooth enough functionsdefined on a half-space. In this setting integration by parts, Leibniz rulesand Holder’s inequality can all be safely applied.

• The technique of partitions of unity and straightening of the boundary.

22.2 Characterizations of W1,p0 (U)

Theorem 22.1. Let U be bounded, open and of class C1. Then for any u ∈W1,p(U) the following are equivalent:

• u can be approximated by elements of C∞c (U) in the W1,p(U) norm.

• The trace of u on ∂U (defined in Lp(∂U)) is zero.

• The extension of u by zero outside U is in W1,p(Rn).

In the proof of this theorem we came across the following fact. For an elementu of W1,p(U), extended by zero outside U , its partial derivative in direction i,defined in the sense of distributions, contains two terms:

• the weak partial derivative in direction i of u, defined in U , and extendedby 0 outside U .

• a boundary term on ∂U stemming from the integration by parts. This isa distribution with support in ∂U .

39

23 Poincare inequality and Dirichlet principle

3 november 2014

23.1 Poincare

Proposition 23.1. Let U ⊆ Rn be open and bounded. There exists C > 0 suchthat for all u ∈ H1

0(U) we have:∫U

|u|2 ≤ C∫U

| gradu|2. (132)

Theorem 23.2. Let U ⊆ Rn be open and bounded. For any f ∈ L2(U) there isa unique u ∈ H1

0(U) such that:

∀u′ ∈ H10(U)

∫gradu · gradu′ =

∫fu′. (133)

Remark 23.1. • Equation (133) is called (by definition) the weak formula-tion of the equation:

−∆u = f. (134)

• The boundary condition u|∂U = 0 is encoded by the condition u ∈ H10(U).

• u is the unique point where the Dirichlet functional reaches its minimumon H1

0(U). Any minimizing sequence converges to u in H10(U).

23.2 Lax-Milgram

The following defines orthogonal projection in a Hilbert space.

Proposition 23.3. Let X be a Hilbert space and Y a closed subspace. Thenfor any u ∈ X, there is a unique v ∈ Y such that u− v ⊥ Y .

Proof. Let 〈·, ·〉 denote the scalar product of X. The vector space Y , equippedwith the restriction of 〈·, ·〉 is a Hilbert space. Also, 〈u, ·〉 is a continuous linearform on Y . We may apply the Riesz representation theorem.

The following theorem, proposition or lemma (according to taste) is knownas Lax-Milgram.

Proposition 23.4. Let X be a Hilbert space, and let a be continuous bilinearform on X, which is coercive in the sense that for some α > 0:

∀u ∈ X a(u, u) ≥ α‖u‖2. (135)

Then for any l ∈ X? there exists a unique u ∈ X such that:

∀v ∈ X a(u, v) = l(v). (136)

Proof. (sketch)Let 〈·, ·〉 denote the scalar product of X.For any u ∈ X denote by Au ∈ X the unique solution to the problem:

∀v ∈ X 〈Au, v〉 = a(u, v), (137)

40

given by the Riesz representation theorem.Then A : X → X is linear and continuous. Moreover:

‖Au‖ ≥ α‖u‖. (138)

One deduces that ranA is closed.If ranA 6= X we may find, by the above corollary, a non-zero u ∈ X such

that u ⊥ ranA. We would get:

α‖u‖2 ≤ a(u, u) = 〈Au, u〉 = 0. (139)

Therefore A is an isomorphism.

Remark 23.2. Consider the hypotheses of the proposition and suppose in addi-tion that a is symmetric. Then a is an inner product on X. It defines a normon X which is equivalent to the original one. Therefore (X, a) is a Hilbert spaceand Lax-Milgram reduces to the Riesz representation theorem.

41

24 Second order elliptic PDEs

5 november 2014

24.1 Variational formulation

[1] §6.1, 6.2.1, 6.2.2.

24.2 Galerkin method

Cea’s lemma.

42

25 The space W1,1 of an interval and extension

10 november 2014

25.1 The space W1,1 of an interval

Proposition 25.1. Let a < b be two reals and consider the interval I =]a, b[.(i) Suppose v ∈ L1(I) and define u : I → R by:

u(x) =

∫ x

a

v(y)dy. (140)

Then:

a. u is continuous.

b. u is differentiable almost everywhere, (in the classical sense) with u′(x) =v(x).

c. v is the weak derivative of u.

(ii) We have:

a. There exists C > 0 such that for all u ∈W1,1(I):

‖u‖L∞(I) ≤ C‖u‖W1,1(I). (141)

b. Any u ∈W1,1(I) coincides a.e. with an element of C(I).

c. If u ∈ W1,1(I) ∩ C(I) and v ∈ L1(I) denotes the weak derivative of u, wehave:

∀x ∈ I u(x) = u(a) +

∫ x

a

v(y)dy. (142)

Proof. (i)a: by the Lebesgue dominated convergence theoremb: by the Lebesgue differentiation theorem.c: We write, for any φ ∈ C∞0 (I):

−∫ b

a

u(x)φ′(x)dx = −∫ b

a

(

∫ x

a

v(y)φ′(x)dy)dx, (143)

= −∫ b

a

∫ b

a

v(y)φ′(x)χ(x, y)dydx, (144)

= −∫ b

a

(

∫ b

y

v(y)φ′(x)dx)dy, (145)

=

∫ b

a

v(y)φ(y)dy, (146)

where χ is defined by:

χ(x, y) =

{1 for y ≤ x,0 for x < y,

(147)

and Fubini was used.

43

(ii)a and b : For u ∈ C1(I) we have:

|u(x)| = |u(y)|+ |∫ y

x

u′(z)dz|, (148)

≤ |u(y)|+∫ b

a

|u′(z)|dz. (149)

Hence:

(b− a)|u(x)| ≤∫ b

a

|u(y)|dy + (b− a)

∫ b

a

|u′(z)|dz. (150)

So:

‖u‖L∞(I) ≤1

b− a‖u‖L1(I) + ‖u′‖L1(I). (151)

For u ∈ W 1,1(I), we may find a sequence un ∈ C1(I), such that un → uin W1,1(I). By the inequality (151) it is Cauchy in C(I). The limit in C(I)must be equal a.e. to the limit in W1,1(I) (why ?). So we may consider thatu ∈ C(I). Then (151) holds for each un and we may pass to the limit n→∞.

c: For u ∈ W 1,1(I) ∩ C(I), we may find a sequence un ∈ C1(I), such thatun → u in W1,1(I). By the preceding results it converges uniformly to u. Foreach x ∈ I:

un(x) = un(a) +

∫ x

a

u′n(y)dy. (152)

Since un converges pointwise to u and u′n converges in L1(I) to v, we may letn→∞ in this identity.

Exercise 25.1. Let I be a bounded open interval. Show that if p ∈]1,∞[, thenfor any u ∈W1,p(I) we have for a.e. x, y ∈ I:

|u(x)− u(y)| ≤ |x− y|1−1/p‖u′‖Lp(I). (153)

Exercise 25.2. Let U ⊆ Rn be open. Pick u ∈ L1(U). Show that for any ε > 0there exists δ > 0 such that for any open V ⊆ U :

meas(V ) ≤ δ ⇒∫V

|u| ≤ ε. (154)

Hint: You may check it first when u is a step function, or a continuous functionwith compact support.

Definition 25.1. A function u : [a, b]→ R is said to be absolutely continuousif for any ε > 0 there exists δ > 0 such that for any n ∈ N, any family of two bytwo disjoint intervals ]ai, bi[⊆ [a, b] for i ∈ [[0, n]], we have:

n∑i=0

|bi − ai| ≤ δ ⇒n∑i=0

|u(bi)− u(ai)| ≤ ε. (155)

Exercise 25.3. Check that functions in W1,1(I) are absolutely continuous, andthat absolutely continuous functions are continuous.

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25.2 The Cantor - Lebesgue function

The Cantor set C is a certain compact subset of [0, 1] with measure 0.The Cantor - Lebesgue function f is a certain continuous function [0, 1] →

[0, 1] with the following two rather paradoxical properties:

• f |C : C → [0, 1] is onto.

• on the complement of C, f is differentiable with differential 0.

In particular the Cantor - Lebesgue function is a continuous and a.e. differ-entiable function, such that the a.e. differential is in L1(]0, 1[), but such that,for x > 0:

f(x) 6= f(0) +

∫ x

0

f ′(y)dy, (156)

where we denote by f ′ the a.e. differential of f (which is 0). Notice that f isnot in W1,1(]0, 1[).

References: page 38 – 39 in [2], or page 37 – 38 in [9].

25.3 Extension

Reference: [1] §5.4.

45

26 Gagliardo-Nirenberg-Sobolev inequality

13 november 2014

Reference: [1] §5.6.1.See also the scanned notes.

46

27 Rellich compactness

18 november 2014

Reference: [1] §5.7.Complements: see also the scanned notes for more information on compact

operators.

47

A Compulsory exercises

To be handed in 29 october 2014.

Exercise A.1. Let u and v be two continuous maps Rn → R. Suppose suppu ⊆E and supp v ⊆ F , with E and F closed subsets of Rn, F being, in addition,bounded.

(a) Show that E + F is closed, where we denote:

E + F = {x+ y : x ∈ E, y ∈ F}. (157)

(b) Show that supp(u ∗ v) ⊆ E + F .(c) (optional) Discuss whether this inclusion is strict or not.

Exercise A.2. Let X be a real Hilbert space, with scalar product 〈·, ·〉. Let Ebe a closed convex subset of X. Suppose that f is a continuous linear form onX. Let J : X → R be the functional defined by:

J(u) =1

2〈u, u〉 − f(u). (158)

(a) Show that there is a unique point u in E such that:

∀v ∈ E J(u) ≤ J(v). (159)

(b) Show that u is uniquely characterized by the properties u ∈ E and:

∀v ∈ E 〈u, v − u〉 − f(v − u) ≥ 0. (160)

(c) Deduce from the above, that given v ∈ X there is a unique point u ∈ E,where ‖u− v‖ is minimal.

Exercise A.3. In this exercise we identify C with R2 in the standard way. Wesay that f : C → C is an entire function if, for each x ∈ C, the following limitexists in C:

f ′(x) = limh→0,h∈C\{0}

f(x+ h)− f(x)

h, (161)

and moreover f ′ : C→ C is a continuous function.(a) Show that f is entire iff f is of class C1 and satisfies the partial differential

equation:

∂1f + i∂2f = 0. (162)

(b) Show that complex polynomials are entire functions.(c) Show that if f is entire and does not take the value 0, then 1/f is entire

and (1/f)′ = −f ′/f2.(d) Show that if f is entire and of class C2, then the real and imaginary parts

of f are harmonic functions.(e) Use Liouville’s theorem and the above, to show that any non–constant

complex polynomial has a root.

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Exercise A.4. (a) On R4 we denote the coordinates of x ∈ R4 by x = (x0, x1, x2, x3),so that the partial derivatives become ∂0, ∂1, ∂2, ∂3. Pick c > 0. The wave equa-tion, for a function u : R4 → R is:

∂20u = c2(∂21 + ∂22 + ∂23)u. (163)

Assume there is smooth function v : R3 → C and ω > 0 such that, for allx ∈ R4:

u(x0, x1, x2, x3) = Re(exp(iωx0)v(x1, x2, x3)). (164)

Show that u solves (163) if and only if v solves the Helmholtz equation on R3:

∆v + k2v = 0. (165)

with k = ω/c.(b) For y ∈ R3 define the plane wave vy : R3 → C by, for all x ∈ R3:

vy(x) = exp(iy · x) (166)

Show that vy satisfies the Helmholtz equation (165) if and only if |y| = k.(c) A smooth spherically symmetric function v : R3 \ {0} → C is one such

that there exists a smooth function w :]0,+∞[→ C such that:

∀x 6= 0 v(x) = w(|x|). (167)

Find all smooth functions w :]0,+∞[→ C such that v defined by (167) solvesthe Helmholtz equation (165) on R3 \ {0}.

Hint: look at the last question first.(d) Show that for all ψ ∈ C∞(R3) and all φ ∈ C∞c (R3):∫

R3

(∆ψ + k2ψ)φ =

∫R3

ψ(∆φ+ k2φ). (168)

(e) Define w :]0,+∞[→ C by:

w(r) =exp(ikr)

4πr, (169)

and v by (167). Check that for any R > 0, w ∈ L1(B(0, R)).Show that for all φ ∈ C∞c (R3):∫

R3

v(∆φ+ k2φ) = −φ(0). (170)

Restate this formula in terms of the Dirac delta.

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B Various exercises

Exercise B.1. Choose a < b in R and let I be the interval I =]a, b[. Show thatthere exists a constant C such that for all u ∈ H1(I) = W1,2(I):

‖u‖L∞(I) ≤ C‖u‖1/2L2(I)‖u‖

1/2H1(I). (171)

Hint: use that for smooth functions:

u(y)2 =

∫ y

x

2u(s)u′(s)ds+ u(x)2. (172)

Exercise B.2. Choose a < b in R and let I be the interval I =]a, b[. Show thatthere exists a C > 0 such that for all u ∈ H1(I):

‖u− 1

b− a

∫I

u‖L2(I) ≤ C‖u′‖L2(I). (173)

Hint: Use the compactness of the injection H1(I) → L2(I), or that smoothfunctions on an interval attain their average.

Exercise B.3. 1. Let I be the interval ]a, b[. Define for any smooth functionu ∈ C∞(I), the affine function PIu : I → R by:

∀x ∈ I (PIu)(x) = u(a) +x− ab− a

(u(b)− u(a)). (174)

Check that PI is a projection. Show that PI has a unique extension to acontinuous operator H1(I)→ H1(I).

2. Deduce from Exercise B.2 that there exists a C > 0 such that for allu ∈ H1(I):

‖u− PIu‖L2(I) ≤ C‖u′‖L2(I). (175)

3. In the preceding question a and b were considered given – we now studythe dependence of the constant C on a and b. Show that there exists aC > 0 such that for all a < b and all u ∈ H1(I) (with I =]a, b[) :

‖u− PIu‖L2(I) ≤ C(b− a)‖u′‖L2(I). (176)

Hint : Use the preceding result in the case I =]0, 1[ and make a change avariable from ]0, 1[ to ]a, b[.

4. (Optional) By similar arguments show that there exists a C > 0 such thatfor any h > 0, any interval I of length h, we have for any u ∈ H2(I):

‖(u− PIu)′‖L2(I) ≤ Ch‖u′′‖L2(I), (177)

and for any u ∈ H1(I):

‖(PIu)′‖L2(I) ≤ C‖u′‖L2(I), (178)

50

Exercise B.4. Let I be the interval ]0, 1[.

1. Pick n > 0 an integer. Define for integer i ∈ [0, n]:

xni = i/n ∈ I. (179)

Define the function λni : I → R by:

λni (x) =

0 for x ≤ xni−1,n(x− xni−1) for xni−1 < x ≤ xni ,1− n(x− xni ) for xni < x ≤ xni+1,0 for xni+1 < x.

(180)

Check that λi is continuous on I, affine on every sub-interval [xnj , xnj+1]

and that:λni (xnj ) = δij . (181)

Define an operator Pn : H1(I)→ H1(I) by:

Pnu =

n∑i=0

u(xi)λni . (182)

Show that Pn is a continuous projector in H1(I).

Show, using (176) and (177) on each subinterval [xnj , xnj+1], that for any

u ∈ H2(I) there exists C > 0 such that for all n:

‖u− Pnu‖H1(I) ≤ C/n. (183)

2. Define Xn by:Xn = span{λni : 0 < i < n}. (184)

Check that for any u ∈ H10(I) we have that Pnu ∈ Xn.

Show using (176) and (178) that Pn is uniformly bounded in H10 (I) →

H1(I), that is, there exists C > 0 such that for all u ∈ H10(I) and all n:

‖Pnu‖H1(I) ≤ C‖u‖H1(I). (185)

Then show that for any u in H10(I), Pnu converges to u in H1(I), as n→∞.

3. Pick δ > 0 and γ ∈ R. For a given f ∈ L2(I) we want to solve theequation:

−δu′′ + γu′ = f. (186)

with boundary conditions u(0) = u(1) = 0.

Define the bilinear form a on H10(I) by, for all u, v ∈ H1

0(I):

a(u, v) = δ

∫u′v′ + γ

∫u′v. (187)

Show that there exists C > 0 such that for all u ∈ H10(I):

a(u, u) ≥ 1/C‖u‖2H1(I). (188)

Apply the Lax-Milgram theorem to show that there is a unique solutionu? ∈ H1

0(I) to the weak formulation of (186). Check that u? ∈ H2(I).

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4. Check that there is a unique un ∈ Xn solving:

∀v ∈ Xn a(un, v) =

∫fv. (189)

5. Show using (188) that there is a constant C such that for all n:

‖u? − un‖H1(I) ≤ C‖u? − Pnu?‖H1(I). (190)

Deduce that un converges to u? in H1(I), at least at the rate O(1/n).

Exercise B.5. (a) Let U ⊆ Rn be open and bounded. Fix p ∈ [1,+∞[. Showthat there exists C > 0 such that for all u ∈W1,p

0 (U):

‖u‖Lp(U) ≤ C‖ gradu‖Lp(U). (191)

(b) Fix u ∈ C∞0 (Rn) and consider for each λ > 0 the function uλ : Rn → Rdefined by:

∀x ∈ Rn uλ(x) = u(λx). (192)

Express ‖uλ‖Lp(Rn) and ‖ graduλ‖Lp(Rn) in terms of λ, ‖u‖Lp(Rn) and ‖ gradu‖Lp(Rn).(c) Let U ⊆ Rn be open. Fix p ∈ [1,+∞[. Suppose that for each R > 0

there exists x ∈ U such that B(x,R) ⊆ U . Show that there does not exist C > 0such that for all u ∈W1,p

0 (U):

‖u‖Lp(U) ≤ C‖ gradu‖Lp(U). (193)

Exercise B.6. Choose p, q ∈ [1,∞[ and θ ∈]0, 1[. Define r by:

1/r = θ/p+ (1− θ)/q. (194)

Let U be an open subset of Rn.a) Show that for all u ∈ Lp(U) ∩ Lq(U), u ∈ Lr(U) and:

‖u‖Lr(U) ≤ ‖u‖θLp(U)‖u‖1−θLq(U). (195)

b) Deduce that if a sequence of functions U → R is bounded in Lp(U) andconverges in Lq(U) then it converges also in Lr(U).

Exercise B.7. Let U =]0, 2π[. For n > 0 define un : U → R by:

un(x) = sin(nx). (196)

a) Show that for all p ∈ [1,∞], the sequence (un)n∈N is bounded in Lp(U).b) Show that for any φ ∈ C0c (U):

limn→∞

∫unφ = 0. (197)

c) Deduce that if a subsequence of (un)n∈N converges in L1(U) to somefunction v, then v = 0.

d) Show that actually no subsequence of (un)n∈N converges in L1(U).e) Is the canonical injection Lp(U)→ L1(U) compact for some p?

52

Exercise B.8. Let U =]0, 1[. For n > 0 define un : U → R by:

un(x) =

{n for x ∈]0, 1/n],0 for x ∈]1/n, 1[.

(198)

a) Remark that un is bounded in L1(U).b) Show that for any φ ∈ C0c (U):

limn→∞

∫unφ = 0. (199)

c) Is there a u ∈ L1(U) such that for all v ∈ L∞(U) we have:

limn→∞

∫unv =

∫uv? (200)

Exercise B.9. Fix an integer n > 0. Let φ : Rn → R be a smooth function suchthat:

suppφ ⊂ B(0, 1), (201)∫φ = 1, (202)

∀x ∈ Rn φ(x) ≥ 0. (203)

∀x ∈ Rn φ(x) = φ(−x). (204)

For any ε > 0 we define φε by:

φε(x) = ε−nφ(x/ε). (205)

a) Pick ψ ∈ C∞c (Rn) and define ψ by ψ(x) = ψ(−x). Show that for allu, v ∈ L2(Rn) we have: ∫

u(v ∗ ψ) =

∫(u ∗ ψ)v. (206)

b) We suppose that we have a function u ∈ L2(Rn) for which there exists aM > 0 such that for each i ∈ {1, · · · , n} and all ε > 0:

‖∂i(φε ∗ u)‖L2(Rn) ≤M. (207)

Show that there exists a C > 0 such that for all ψ ∈ C∞c (Rn):

|∫u∂iψ| ≤ C‖ψ‖L2(Rn). (208)

c) Deduce that u ∈ H1(Rn).

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References

[1] L. C. Evans. Partial differential equations, volume 19 of Graduate Studiesin Mathematics. American Mathematical Society, Providence, RI, secondedition, 2010.

[2] G. B. Folland. Real analysis. Pure and Applied Mathematics (New York).John Wiley & Sons Inc., New York, second edition, 1999. Modern tech-niques and their applications, A Wiley-Interscience Publication.

[3] R. Godement. Analyse mathematique. II. Springer-Verlag, Berlin, secondedition, 2003. Calcul differentiel et integral, series de Fourier, fonctionsholomorphes. [Differential and integral calculus, Fourier series, holomorphicfunctions].

[4] L. Hormander. The analysis of linear partial differential operators. I. Clas-sics in Mathematics. Springer-Verlag, Berlin, 2003. Distribution theory andFourier analysis, Reprint of the second (1990) edition [Springer, Berlin;MR1065993 (91m:35001a)].

[5] S. Lang. Undergraduate analysis. Undergraduate Texts in Mathematics.Springer-Verlag, New York, second edition, 1997.

[6] P. D. Lax. Functional analysis. Pure and Applied Mathematics (New York).Wiley-Interscience [John Wiley & Sons], New York, 2002.

[7] B. P. Rynne and M. A. Youngson. Linear functional analysis. SpringerUndergraduate Mathematics Series. Springer-Verlag London, Ltd., London,second edition, 2008.

[8] E. M. Stein and R. Shakarchi. Fourier analysis. Princeton Lectures inAnalysis, I. Princeton University Press, Princeton, NJ, 2003. An introduc-tion.

[9] E. M. Stein and R. Shakarchi. Real analysis. Princeton Lectures in Analy-sis, III. Princeton University Press, Princeton, NJ, 2005. Measure theory,integration, and Hilbert spaces.

[10] T. Tao. Analysis. II, volume 38 of Texts and Readings in Mathematics.Hindustan Book Agency, New Delhi, second edition, 2009.

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