a gentle introduction to the kaczmarz algorithmorion.math.iastate.edu › esweber › math610 ›...

A Gentle Introduction to the Kaczmarz Algorithm

Eric Weber

Iowa State University

Mathematics ColloquiumDePaul UniversityOctober 27, 2017

Eric Weber A Gentle Introduction to the Kaczmarz Algorithm 1 / 39

Systems of Linear Equations

Q: How do we solve a system of equations? (We assume consistency here).

a11x1 + a12x2 + · · ·+ a1nxn = y1

a21x1 + a22x2 + · · ·+ a2nxn = y2

......

am1x1 + am2x2 + · · ·+ amnxn = ym

Solutions

A: 1) Gaussian Elimination and Backsubstitution:a11 a12 . . . a1n y1a21 a22 . . . a2n y2

......

. . ....

...am1 am2 . . . amn ym

2) Matrix Inversion (m = n; detA 6= 0):

A~x = ~y ⇒ ~x = A−1~y .

3) Moore-Penrose Inversion (m ≥ n; nullity(A) = 0):

~x = A†~y = (ATA)−1AT~y

Solutions

......

. . ....

...am1 am2 . . . amn ym

A~x = ~y ⇒ ~x = A−1~y .

~x = A†~y = (ATA)−1AT~y

Solutions

......

. . ....

...am1 am2 . . . amn ym

A~x = ~y ⇒ ~x = A−1~y .

~x = A†~y = (ATA)−1AT~y

(Non)-Solutions

4) Least-Squares (no solution; Moore-Penrose or gradient-descent )

Find x1, . . . , xn that minimizesm∑j=1

|(aj1x1 + · · ·+ ajnxn)− yj |2

5) Compressed Sensing (many solutions; m < n)

Find the solution x1, . . . , xn that minimizes eithern∑

|xk |0 orn∑

|xk |1

Genome mapping; MRI machines.

6) Kaczmarz Algorithm

(coming soon! these last two form a 1-2 punch in data science)

(Non)-Solutions

|(aj1x1 + · · ·+ ajnxn)− yj |2

|xk |0 orn∑

|xk |1

(Non)-Solutions

|(aj1x1 + · · ·+ ajnxn)− yj |2

|xk |0 orn∑

|xk |1

Compressed Sensing, An Example

Known:

{f (t0), f (t1), . . . , f (tN)} (Samples)

f (x) = a0 + a1x + · · ·+ aNxN

Unknown:

f , i.e. a0, a1, . . . , aN

(N + 1 unknown variables in N + 1 dimensions)

Can we recover/reconstruct f (x)? Yes:1 t0 t20 . . . tN01 t1 t21 . . . tN1...

......

. . ....

1 tN t2N . . . tNN

a0a1...aN

f (t0)f (t1)

...f (tN)

We need N + 1 samples or “measurements” to uniquely determine the vector f inN + 1 dimensions. Here, any N + 1 distinct samples will work.

Known:

{f (t0), f (t1), . . . , f (tN)} (Samples)

f (x) = a0 + a1x + · · ·+ aNxN

Unknown:

f , i.e. a0, a1, . . . , aN

......

. . ....

1 tN t2N . . . tNN

a0a1...aN

f (t0)f (t1)

...f (tN)

Known:

{f (t0), f (t1), . . . , f (tN)} (Samples)

f (x) = a0 + a1x + · · ·+ aNxN

Unknown:

f , i.e. a0, a1, . . . , aN

......

. . ....

1 tN t2N . . . tNN

a0a1...aN

f (t0)f (t1)

...f (tN)

Known:

{f (t0), f (t1), . . . , f (tN)} (Samples)

f (x) = a0 + a1x + · · ·+ aNxN

Unknown:

f , i.e. a0, a1, . . . , aN

......

. . ....

1 tN t2N . . . tNN

a0a1...aN

f (t0)f (t1)

...f (tN)

Compressed Sensing, An Example (cont’d)

Known:

{f (t0), f (t1), . . . , f (t2N), f (t2N+1)} (Samples)

f (x) = a0xn0 + a1x

n1 + · · ·+ aNxnN

Unknown:

f , i.e. a0, a1, . . . , aN AND n0, n1, . . . , nN

(2N + 2 unknown variables in infinite dimensions!)

Can we recover/reconstruct f (x)? Yes: (if we are careful!)

We choose t0 > 0, t0 6= 1, then tj = t j+10 .

Proof.

Variation of Prony’s algorithm (1795)!

Known:

f (x) = a0xn0 + a1x

n1 + · · ·+ aNxnN

Unknown:

f , i.e. a0, a1, . . . , aN AND n0, n1, . . . , nN

Proof.

Known:

f (x) = a0xn0 + a1x

n1 + · · ·+ aNxnN

Unknown:

f , i.e. a0, a1, . . . , aN AND n0, n1, . . . , nN

Proof.

Known:

f (x) = a0xn0 + a1x

n1 + · · ·+ aNxnN

Unknown:

f , i.e. a0, a1, . . . , aN AND n0, n1, . . . , nN

Proof.

Reconstruction

Back to systems of equations:we write ~aj = (aj1, aj2, . . . , ajn), ~x = (x1, x2, . . . , xn), and

~x · ~a1 = y1

~x · ~am = ym

So, the data we have are the dot products of ~x with the ~aj ’s (linearmeasurements), and we want to recover ~x from that data.

Reconstruction

Back to systems of equations:we write ~aj = (aj1, aj2, . . . , ajn), ~x = (x1, x2, . . . , xn), and

~x · ~a1 = y1

~x · ~am = ym

So, the data we have are the dot products of ~x with the ~aj ’s (linearmeasurements), and we want to recover ~x from that data.

Reconstruction (cont’d)

If {~a1, . . . , ~an} is an orthonormal basis (think ~i , ~j , ~k), then

~x = (~x · ~a1)~a1 + · · ·+ (~x · ~an)~an = y1~a1 + · · ·+ yn~an.

(Note the form of the terms–projections!)

If {~a1, . . . , ~am} is a basis (or even just a spanning set–i.e. frame), then there exists

{~b1, . . . , ~bm} such that

~x = (~x · ~a1)~b1 + · · ·+ (~x · ~am)~bm = y1~b1 + · · ·+ ym~bm.

Calculating ~bj ’s is easy on paper and hard on a computer!

{~b1, . . . , ~bm} such that

Kaczmarz Algorithm

Given {~an}∞n=0 ⊂ H (unit vectors) and ~x · ~an, can we recover ~x? Note: yes ifONB/frame.

~x0 = (~x · ~a0)~a0

~xn = ~xn−1 + ((~x − ~xn−1) · ~an)~an

If limn→∞‖~x − ~xn‖ = 0 for all ~x , then the sequence {~an}∞n=0 is said to be effective.

Theorem (Kaczmarz, 1937)

Suppose {~a0, . . . , ~am−1} is a spanning set of unit vectors for H, and define thesequence by periodizing:

~am = ~a0, ~am+1 = ~a1, . . .

then {~an}∞n=0 is effective.

Kaczmarz Algorithm

Given {~an}∞n=0 ⊂ H (unit vectors) and ~x · ~an, can we recover ~x? Note: yes ifONB/frame.

~x0 = (~x · ~a0)~a0

~xn = ~xn−1 + ((~x − ~xn−1) · ~an)~an

If limn→∞‖~x − ~xn‖ = 0 for all ~x , then the sequence {~an}∞n=0 is said to be effective.

Theorem (Kaczmarz, 1937)

Suppose {~a0, . . . , ~am−1} is a spanning set of unit vectors for H, and define thesequence by periodizing:

~am = ~a0, ~am+1 = ~a1, . . .

then {~an}∞n=0 is effective.

Proof of the Kaczmarz Theorem

We want to show that ‖~x − ~xn‖ → 0. Calculate:

~x − ~xn = (~x − ~xn−1)− ((~x − ~xn−1) · ~an)~an

Note that the second term is a projection.

~x − ~xn = (I − Pn)(~x − ~xn−1)

= (I − Pn)(I − Pn−1) . . . (I − P1)(I − P0)~x

Since projections only decrease the norm, we have ‖~x − ~xn‖ ≤ ‖~x‖. Note that:

~x − ~x2m = (I − Pm−1) . . . (I − P1)(I − P0)(I − Pm−1) . . . (I − P1)(I − P0)~x

so if ‖~x − ~xm‖ = α‖~x‖, we have ‖~x − ~x2m‖ = α2‖~x‖.

~x − ~xn = (~x − ~xn−1)− ((~x − ~xn−1) · ~an)~an

Note that the second term is a projection. Thus,

~x − ~xn = (I − Pn)(~x − ~xn−1)

= (I − Pn)(I − Pn−1) . . . (I − P1)(I − P0)~x

~x − ~xn = (~x − ~xn−1)− ((~x − ~xn−1) · ~an)~an

Note that the second term is a projection. Thus,

~x − ~xn = (I − Pn)(~x − ~xn−1)

= (I − Pn)(I − Pn−1) . . . (I − P1)(I − P0)~x

Proof of the Kaczmarz Theorem (cont’d)

Again, since we have projections, α ≤ 1. If α < 1 for every ~x we are done. So,can α = 1?

What if it were? We use the Rado-Horn Theorem:

α ≤ ‖(I − Pm−1) . . . (I − P1)(I − P0)‖ ≤ 1

and if the norm is 1, then there exists an eigenvector ~y with eigenvalue 1. Thus:

~y = (I − Pm−1) . . . (I − P1)(I − P0)~y

= (I − Pm−1) . . . (I − P1)~y

= (I − Pm−1)~y .

Thus, ~y · ~aj = 0 for all j , so we must have ~y = 0!

Again, since we have projections, α ≤ 1. If α < 1 for every ~x we are done. So,can α = 1?What if it were? We use the Rado-Horn Theorem:

α ≤ ‖(I − Pm−1) . . . (I − P1)(I − P0)‖ ≤ 1

~y = (I − Pm−1) . . . (I − P1)(I − P0)~y

= (I − Pm−1) . . . (I − P1)~y

= (I − Pm−1)~y .

α ≤ ‖(I − Pm−1) . . . (I − P1)(I − P0)‖ ≤ 1

and if the norm is 1, then there exists an eigenvector ~y with eigenvalue 1.

~y = (I − Pm−1) . . . (I − P1)(I − P0)~y

= (I − Pm−1) . . . (I − P1)~y

= (I − Pm−1)~y .

α ≤ ‖(I − Pm−1) . . . (I − P1)(I − P0)‖ ≤ 1

~y = (I − Pm−1) . . . (I − P1)(I − P0)~y

= (I − Pm−1) . . . (I − P1)~y

= (I − Pm−1)~y .

So, Why Kaczmarz?

Advantages:

1 Fast! Low complexity, geometric convergence.

2 easy to program

3 simple assumptions on the vectors

Disadvantages:

1 non-exact solution: at best an approximate solution (not a problemnumerically)

2 slower/more complex IF we know a priori an alternative reconstruction (dualbasis/frame)

So, Why Kaczmarz?

Advantages:

1 Fast! Low complexity, geometric convergence.

2 easy to program

3 simple assumptions on the vectors

Disadvantages:

1 non-exact solution: at best an approximate solution (not a problemnumerically)

2 slower/more complex IF we know a priori an alternative reconstruction (dualbasis/frame)

The Kaczmarz Algorithm and Compressed Sensing

Suppose we have an underdetermined system A~x = ~y (the matrix A is short andfat). We want to find a solution ~x with the fewest non-zero entries.

Let us denote the columns of A by ~A1, . . . , ~An.

Goal: find a subset ~Ak1 , . . . ,~Akq of the columns of A with the fewest elements

such that ~y is in the span of those column vectors.

This is a combinatorial search–extremely slow (NP Hard)!

The main results in compressed sensing say roughly the following: if A has acertain property (hard to get, but possible) then the sparsest solution ( minimizes∑n

k=1 |xk |0 ) is the one that minimizes∑n

k=1 |xk |1.

The Kaczmarz Algorithm and Compressed Sensing (cont’d)

We assume a priori that there exists q columns of A whose span contains ~y ; wejust don’t know which ones.

Alternative approach:

Step 1: Choose ~Ak1 , . . . ,~Ak2q using an estimate on which columns are the

correct choice;

Step 2: Apply Kaczmarz to the system(~Ak1

~Ak2 . . .~Ak2q

)~z = ~y

to obtain an estimate ~z to ~x ;

Step 3: Use ~z to get a better estimate on which columns to choose, and repeatStep 2.

correct choice;

~Ak2 . . .~Ak2q

)~z = ~y

correct choice;

~Ak2 . . .~Ak2q

)~z = ~y

correct choice;

~Ak2 . . .~Ak2q

)~z = ~y

Kaczmarz Algorithm in Infinite Dimensions

Theorem (Kwapien & Mycielski, 2001)

If {φn}∞n=1 ⊂ H is a stationary sequence with dense span, then it is an effectivesequence if and only if it’s spectral measure is either Lebesgue measure or purelysingular.

Fourier Series

Theorem (Herr & W., 2015)

If µ is a singular Borel probability measure on (−1/2, 1/2), then the sequence{e2πinx

}∞n=0

is effective in L2(µ). As a consequence, any element f ∈ L2(µ)possesses a Fourier series

f (x) =∞∑n=0

cne2πinx ,

where the sum converges in the L2(µ) norm.

The Fourier coefficients cn are givenby

∫ 1/2

−1/2f (x)gn(x) dµ(x),

where {gn}∞n=0 is the auxiliary sequence of{e2πinx

}∞n=0

in L2(µ).

Fourier Series

Theorem (Herr & W., 2015)

If µ is a singular Borel probability measure on (−1/2, 1/2), then the sequence{e2πinx

}∞n=0

is effective in L2(µ). As a consequence, any element f ∈ L2(µ)possesses a Fourier series

f (x) =∞∑n=0

cne2πinx ,

where the sum converges in the L2(µ) norm. The Fourier coefficients cn are givenby

∫ 1/2

−1/2f (x)gn(x) dµ(x),

where {gn}∞n=0 is the auxiliary sequence of{e2πinx

}∞n=0

in L2(µ).

Inversion Lemma

Lemma (Herr & W., 2015)

There exists a sequence {αn}∞n=0 such that

gn(x) =n∑

αn−je2πijx .

The sequence is given by

µ+(z)=∞∑n=0

µ+(z) =

∫ 1/2

−1/2

1− e−2πitzdµ(t).

Inversion Lemma

Lemma (Herr & W., 2015)

There exists a sequence {αn}∞n=0 such that

gn(x) =n∑

αn−je2πijx .

The sequence is given by

µ+(z)=∞∑n=0

µ+(z) =

∫ 1/2

−1/2

1− e−2πitzdµ(t).

The Paley-Wiener Space

We can reconstruct an unknown vector ~x from its “samples” (polynomials) or“linear measurements” (A~x = ~y) quite well in finite dimensions. However, thesampling algorithm usually occurs in infinite dimensions! i.e. the vector space ofcontinuous functions on [0, 1].

Consider the vector space of entire functions f : C→ C that satisfies these twoconditions:

1∫∞−∞ |f (t)|2dt is finite;

2 |f (z)| ≤ Ceπ|z| for some constant C .

This is the Paley-Wiener space.

The Paley-Wiener Space

We can reconstruct an unknown vector ~x from its “samples” (polynomials) or“linear measurements” (A~x = ~y) quite well in finite dimensions. However, thesampling algorithm usually occurs in infinite dimensions! i.e. the vector space ofcontinuous functions on [0, 1].

Consider the vector space of entire functions f : C→ C that satisfies these twoconditions:

1∫∞−∞ |f (t)|2dt is finite;

2 |f (z)| ≤ Ceπ|z| for some constant C .

This is the Paley-Wiener space.

The Shannon Sampling Theorem

Theorem (Shannon-Whittaker-Kotelnikov ( ∼1945, ∼1915, ∼1935))

If f is in the Paley-Wiener space, then

f (x) =∑n∈Z

(sin(π(x − n)

π(x − n)

In other words: Known:

1 f is in the Paley-Wiener space;

2 f (n) for all n ∈ Zthen we can recover f from this data.

This is the foundation of digital communications: cellphones, digital audio(compact discs, MP3, etc), CAT scans, MRI’s, . . . (Nyquist rate; analog to digitalconversion)

f (x) =∑n∈Z

(sin(π(x − n)

π(x − n)

f (x) =∑n∈Z

(sin(π(x − n)

π(x − n)

Paley-Wiener Theorem

Theorem (Paley-Wiener Theorem (1933))

If f is in the Paley-Wiener space, then there exists a function g ∈ L2(−1/2, 1/2)such that

f (z) =

∫ 1/2

−1/2g(t)e−2πitzdt.

Equivalently:If f satisfies the following conditions:

1 f is entire;

n∈Z |f (n)|2 is finite;

f (x) =∑n∈Z

(sin(π(x − n)

π(x − n)

then f is the Fourier transform of some g .

Paley-Wiener Theorem

Theorem (Paley-Wiener Theorem (1933))

If f is in the Paley-Wiener space, then there exists a function g ∈ L2(−1/2, 1/2)such that

f (z) =

∫ 1/2

−1/2g(t)e−2πitzdt.

Equivalently:If f satisfies the following conditions:

1 f is entire;

n∈Z |f (n)|2 is finite;

f (x) =∑n∈Z

(sin(π(x − n)

π(x − n)

then f is the Fourier transform of some g .

Singular Measures

Fix a singular measure µ on (−1/2, 1/2). Question: when can an entire functionF be written as

F (z) =

∫ 1/2

−1/2f (t)e−2πitzdµ(t)

for some f ∈ L2(µ)?

Answer: Kaczmarz Algorithm!

Using the Kaczmarz algorithm, we are able to provide complete characterizationsusing i) a sampling theory idea and ii) an interpolation idea.

Singular Measures

Fix a singular measure µ on (−1/2, 1/2). Question: when can an entire functionF be written as

F (z) =

∫ 1/2

−1/2f (t)e−2πitzdµ(t)

for some f ∈ L2(µ)?

Answer: Kaczmarz Algorithm!

Using the Kaczmarz algorithm, we are able to provide complete characterizationsusing i) a sampling theory idea and ii) an interpolation idea.

The Paley-Wiener Theorem for µ

Theorem (W., 2017)

Let µ be a singular Borel probability measure on (−1/2, 1/2). Let {αn}∞n=0 be thesequence of scalars induced by µ by the Inversion Lemma. The entire function Fhas the form

F (z) =

∫ 1/2

−1/2f (x)e−2πitz dµ(t)

for some f ∈ L2(µ) if and only if F satisfies

∞∑n=0

∣∣∣∣∣∣n∑

αn−jF (j)

∣∣∣∣∣∣2

2 for all z ∈ C,

F (z) =∞∑n=0

n∑j=0

αn−jF (j)

( n∑l=0

αn−l µ(z − l)

Theorem (W., 2017)

F (z) =

∫ 1/2

∞∑n=0

∣∣∣∣∣∣n∑

αn−jF (j)

∣∣∣∣∣∣2

2 for all z ∈ C,

F (z) =∞∑n=0

n∑j=0

αn−jF (j)

( n∑l=0

αn−l µ(z − l)

Theorem (W., 2017)

F (z) =

∫ 1/2

∞∑n=0

∣∣∣∣∣∣n∑

αn−jF (j)

∣∣∣∣∣∣2

2 for all z ∈ C,

F (z) =∞∑n=0

n∑j=0

αn−jF (j)

( n∑l=0

αn−l µ(z − l)

Necessity: Apply Fourier transform to

f =∞∑n=0

〈f , gn〉gn.

Sufficiency: Define f ∈ L2(µ) by

f =∞∑n=0

n∑j=0

αn−jF (j)

then apply Fourier transform to obtain

f (z) = F (z).

Necessity: Apply Fourier transform to

f =∞∑n=0

〈f , gn〉gn.

Sufficiency: Define f ∈ L2(µ) by

f =∞∑n=0

n∑j=0

αn−jF (j)

then apply Fourier transform to obtain

f (z) = F (z).

Theorem (W. 2017)

Suppose µ is a singular Borel probability measure on (−1/2, 1/2), and let b be theinner function associated to µ by the Herglotz Representation. The entire functionF is the Fourier transform f for some f ∈ L2(µ) if and only if

(i) |F (z)| ≤ ε(|z |)eπ|z| with ε(r) = o(1);

(ii) the following inclusions hold:

G+(z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b), G−(z) :=

∑∞n=0 F (−n)zn

µ+(z)∈ H(b);

(iii) the L2(µ)-boundaries of G+ and G− satisfy the relationship

G?+ = G?−.

Theorem (W. 2017)

G+(z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b), G−(z) :=

∑∞n=0 F (−n)zn

µ+(z)∈ H(b);

G?+ = G?−.

Theorem (W. 2017)

G+(z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b), G−(z) :=

∑∞n=0 F (−n)zn

µ+(z)∈ H(b);

G?+ = G?−.

Theorem (W. 2017)

G+(z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b), G−(z) :=

∑∞n=0 F (−n)zn

µ+(z)∈ H(b);

G?+ = G?−.

A No-Go Result

Denote: PW (µ) = {f |f ∈ L2(µ)}, Eτ the collection of all entire functions ofexponential type at most τ .

Theorem

Suppose PW (µ) = Cτ ∩ L2(w) for some τ ∈ (0, π] and some weight or measure won R with ‖f ‖µ ' ‖f ‖w . Then there exists a Riesz basis of the form

{ωne2πiλnx}n∈Z ⊂ L2(µ) (1)

for some sequence {λn} ⊂ R and ωn > 0.

A No-Go Result

Denote: PW (µ) = {f |f ∈ L2(µ)}, Eτ the collection of all entire functions ofexponential type at most τ .

Theorem

Suppose PW (µ) = Cτ ∩ L2(w) for some τ ∈ (0, π] and some weight or measure won R with ‖f ‖µ ' ‖f ‖w . Then there exists a Riesz basis of the form

{ωne2πiλnx}n∈Z ⊂ L2(µ) (1)

for some sequence {λn} ⊂ R and ωn > 0.

Herglotz Representation Theorem and the space H(b)

Theorem

There is a 1-to-1 correspondence between the nonconstant inner functions b in H2

and the nonnegative singular Borel measures µ on T ≡ [0, 1) given by

(1 + b(z)

1− b(z)

1− |z |2

|ξ − z |2dµ(ξ).

We will say that b corresponds to µ, and that µ corresponds to b. Theconstruction of the de Branges-Rovnyak space H(b) is based on Toeplitzoperators, but here suffice it to say that for b an inner function, we have

H(b) = H2 bH2.

Normalized Cauchy Transform

Given a measure µ on (−1/2, 1/2), the normalized Cauchy transform is theoperator Vµ from L2(µ) to the set of functions on C \ T given by

Vµf (z) =

∫ 1/2

−1/2

1− ze−2πixdµ(x)∫ 1/2

−1/2

1− ze−2πixdµ(x)

Clark showed that if µ is a singular Borel probability measure and b is itscorresponding inner function, then Vµ maps L2(µ) unitarily onto H(b).

Re-Expression of the Normalized Cauchy Transform

Theorem (H.&W., 2015)

Let µ be a singular Borel probability measure, and {gn}∞n=0 the auxiliary sequenceof {en}∞n=0 in L2(µ). Then for f ∈ L2(µ),

Vµf (z) =∞∑n=0

〈f , gn〉µ zn.

Thus, every function F ∈ H(b) is of the form F (z) =∑∞

n=0 〈f , gn〉µ zn. Since

f =∑∞

n=0 〈f , gn〉µ e2πinx and Fr (x) :=∑∞

n=0 〈f , gn〉µ re2πinx , Abel summability

shows that limr→1− ‖Fr − f ‖µ = 0, and so f is an L2(µ) boundary function of F .

Re-Expression of the Normalized Cauchy Transform

Theorem (H.&W., 2015)

Let µ be a singular Borel probability measure, and {gn}∞n=0 the auxiliary sequenceof {en}∞n=0 in L2(µ). Then for f ∈ L2(µ),

Vµf (z) =∞∑n=0

〈f , gn〉µ zn.

Thus, every function F ∈ H(b) is of the form F (z) =∑∞

n=0 〈f , gn〉µ zn. Since

f =∑∞

n=0 〈f , gn〉µ e2πinx and Fr (x) :=∑∞

n=0 〈f , gn〉µ re2πinx , Abel summability

shows that limr→1− ‖Fr − f ‖µ = 0, and so f is an L2(µ) boundary function of F .

The Paley-Wiener Theorem for µ, redux

For an entire function F of exponential type, we use hF to denote thePhragmen-Lindelof indicator function.

Theorem (W. 2017)

Suppose µ is a singular Borel probability measure with support in[α, β] ⊂ [−1/2, 1/2] where β − α < 1. Let b be the inner function associated to µvia the Herglotz Representation. The entire function F is the Fourier transform ffor some f ∈ L2(µ) if and only if

(i) F is of exponential type;

(ii) the indicator function of F satisfies hF (π

2) ≤ 2πβ and hF (−π

2) ≤ −2πα;

(iii) the following inclusion holds:

GF (z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b)

i.e. the function GF is in the kernel of the Toeplitz operator Tb.

Theorem (W. 2017)

2) ≤ −2πα;

GF (z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b)

Theorem (W. 2017)

2) ≤ −2πα;

GF (z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b)

Theorem (W. 2017)

2) ≤ −2πα;

GF (z) :=

∑∞n=0 F (n)zn

µ+(z)∈ H(b)

An Interpolation Problem

Suppose µ is a singular Borel probability measure on T, b is the inner function onD associated to µ via the Herglotz representation, and suppose {an}∞n=0 ⊂ C. Thefollowing conditions are equivalent:

(i) there exists a function f ∈ L2(µ) with the property that

∫Tf (x)e−2πinx dµ(x);

(ii) the following inclusion holds:

Ga(z) :=

∑∞n=0 anz

µ+(z)∈ H(b).

An Interpolation Problem

Suppose µ is a singular Borel probability measure on T, b is the inner function onD associated to µ via the Herglotz representation, and suppose {an}∞n=0 ⊂ C. Thefollowing conditions are equivalent:

(i) there exists a function f ∈ L2(µ) with the property that

∫Tf (x)e−2πinx dµ(x); (2)

(ii) the following inclusion holds:

Ga(z) :=

∑∞n=0 anz

µ+(z)∈ H(b).

Outline of Proof

Version using hF :

1 Condition (iii) says that {F (n)} can be interpolated by some f , soF (n) = f (n) for n ∈ N0;

2 Condition (i) and (ii) says that F (z) = f (z) for all z by Carlson’s theorem.

Version using growth condition is similar, but Carlson’s theorem does not apply.We use a generalization of Carlson’s theorem from Boas’ book.

Outline of Proof

Version using hF :

Outline of Proof

Version using hF :

The End

Thank you!

Selected Works Cited

The Paley-Wiener Theorem

An entire function F can be written in the form

F (z) =

∫ 1/2

−1/2f (t)e−2πitzdt

if and only if F satisfies:

1 F is of exponential type π;

2 F (t) ∈ L2(R).

}Eπ ∩ L2(R)

Alternative characterization: F satisfies

1 ∑n∈Z|F (n)|2 <∞;

F (z) =∑n∈Z

F (n)sinc(z − n).

The Paley-Wiener Theorem

An entire function F can be written in the form

F (z) =

∫ 1/2

−1/2f (t)e−2πitzdt

if and only if F satisfies:

1 F is of exponential type π;

2 F (t) ∈ L2(R).

}Eπ ∩ L2(R)

Alternative characterization: F satisfies

1 ∑n∈Z|F (n)|2 <∞;

F (z) =∑n∈Z

F (n)sinc(z − n).

A Shannon Sampling Formula

Theorem (H. & W., 2015)

Let µ be a singular Borel probability measure on (−1/2, 1/2). Let {αn}∞n=0 be thesequence of scalars induced by µ by the Inversion Lemma. Suppose F : R→ C isof the form

F (y) =

∫ 1/2

−1/2f (x)e−2πiyx dµ(x)

for some f ∈ L2(µ). Then

F (y) =∞∑n=0

n∑j=0

αn−jF (j)

µ(y − n),

where the series converges uniformly in y .

A Shannon Sampling Formula

Theorem (H. & W., 2015)

F (y) =

∫ 1/2

F (y) =∞∑n=0

n∑j=0

αn−jF (j)

µ(y − n),

An Alternative Sampling Formula

Since the {gn} form a Parseval frame, we obtain the following variation.

Theorem (H. & W., 2015)

F (y) =

∫ 1/2

F (y) =∞∑n=0

n∑j=0

αn−jF (j)

( n∑l=0

αn−l µ(y − l)

An Alternative Sampling Formula

Since the {gn} form a Parseval frame, we obtain the following variation.

Theorem (H. & W., 2015)

F (y) =

∫ 1/2

F (y) =∞∑n=0

n∑j=0

αn−jF (j)

( n∑l=0

αn−l µ(y − l)

Fourier Series without Frames

The Paley-Wiener Theorem via a Sampling Criteria

The Paley-Wiener Theorem via an Interpolation Criteria

a gentle introduction to the kaczmarz algorithmorion.math.iastate.edu › esweber › math610 ›...

Documents