Folland: Real Analysis, Chapter 8Sebastien Picard

Problem 8.3Let η(t) = e−1/t for t > 0, η(t) = 0 for t ≤ 0.

a. For k ∈ N and t > 0, η(k)(t) = Pk(1/t)e−1/t where Pk is a polynomial of degree 2k.

b. η(k)(0) exists and equals zero for all k ∈ N.

Solution:(a) It is clear that for t > 0 all derivatives η(k)(t) exist and can by computed by the usual rules ofcalculus. We proceed by induction. First, notice

η(1)(t) =d

dte−1/t =

1

t2e−1/t = P1(1/t)e

−1/t,

where P1 = x2 is a polynomial of degree 2.

Suppose η(k)(t) = Pk(1/t)e−1/t for some k, where Pk is a polynomial of degree 2k. Then

η(k+1)(t) =

(

d

dtPk(1/t)

)

e−1/t +Pk(1/t)

t2e−1/t =

(

1

t2Pk(1/t)−

1

t2P ′k(1/t)

)

e−1/t.

This yields η(k+1)(t) = Pk+1(1/t)e−1/t where

Pk+1 = x2Pk − x2P ′k.

Since Pk is a polynomial of degree 2k, x2Pk has highest order coefficient of degree 2k + 2. On theother hand, P ′

k is a polynomial of degree less than or equal to 2k − 1; therefore x2P ′k has degree less

than or equal to 2k + 1. It follows that Pk+1 = x2Pk − x2P ′k is a polynomial of degree 2(k + 1). The

proof follows by induction.

(b) We proceed again by induction. First, we compute

limt→0+

η(t)− η(0)

t= lim

t→0+

e−1/t

t= lim

s→∞se−s = 0.

The left-handed limit is easy since η(t) ≡ 0 when t ≤ 0:

limt→0−

η(t)− η(0)

t= 0.

Therefore, η′(0) exists and is equal to zero.

Suppose η(k)(0) exists and equals zero. We compute η(k+1)(0). Since η(t) ≡ 0 for t ≤ 0, wehave η(m)(t) = 0 for all t < 0 and all m ∈ N. By induction hypothesis, η(k)(0) = 0. Therefore we havethe following left-handed limit:

1

limt→0−

η(k)(t)− η(k)(0)

t= 0.

On the other hand, we can use (a) and the fact that e−x decays faster than any polynomial tocompute the following right-handed limit:

limt→0+

η(k)(t)− η(k)(0)

t= lim

t→0+

Pk(1/t)e−1/t

t= lim

s→∞sPk(s)e

−s = 0.

This shows that η(n+1)(0) = 0. By induction, η(k)(0) exists and equals zero for all k ∈ N.

Problem 8.4If f ∈ L∞ and ||τyf − f ||∞ → 0 as y → 0, then f agrees a.e. with a uniformly continuous function.(Let Arf be as in Theorem 3.18. Then Arf is uniformly continuous for r > 0 and uniformly Cauchyas r → 0.)

Solution:Define Arf as in Theorem 3.18:

Arf(x) =1

m(B(r, x))

∫

B(r,x)

f(z)dz.

The first step is to show that for r > 0, Arf is uniformly continuous. Choose δ > 0 such that forall c ∈ R

n such that |c| < δ, we have ||τcf − f ||∞ < ǫ. Then when |x− y| < δ, we have

|Arf(x)− Arf(y)| =1

m(B(r, x))

∣

∣

∣

∣

∫

Br(x)

f(z)dz −

∫

Br(y)

f(z)dz

∣

∣

∣

∣

=1

m(B(r, x))

∣

∣

∣

∣

∫

Br(x)

f(z)dz −

∫

Br(x)

f(z + y − x)dz

∣

∣

∣

∣

≤1

m(B(r, x))

∫

Br(x)

|f(z)− f(z − (x− y))|dz

≤||τx−yf − f ||∞m(B(r, x))

∫

Br(x)

dz

< ǫ.

Next, we show that Arf is uniformly Cauchy. Choose δ > 0 such that ||τcf − f ||∞ < ǫ/2 when|c| < δ. Then for all r, s > 0 such that r < δ, s < δ, for any x we have

2

|Arf(x)−Asf(x)| ≤ |Arf(x)− f(x)|+ |Asf(x)− f(x)|

=1

m(B(r, x))

∣

∣

∣

∣

∫

Br(x)

(f(z)− f(x))dz

∣

∣

∣

∣

+1

m(B(s, x))

∣

∣

∣

∣

∫

Bs(x)

(f(z)− f(x))dz

∣

∣

∣

∣

≤1

m(B(r, x))

∫

Br(x)

|τz−xf(z)− f(z)|dz +1

m(B(s, x))

∫

Bs(x)

|τz−xf(z)− f(z)|dz

< ǫ/2 + ǫ/2 = ǫ.

Taking the supremum over all points x, we see that Arf is uniformly Cauchy.

At each point x, {Arf(x)} is Cauchy, so we can define f(x) = limr→0Arf(x). We show Arfconverges to f as r → 0 in the uniform norm. Select δ > 0 such that ||As1f − As2f ||u < ǫ for all0 < s1 < δ, 0 < s2 < δ. Then if 0 < r < δ, for each point x we have

|Arf(x)− f(x)| = lims→0

|Arf(x)− Asf(x)| ≤ lims→0

||Arf −Asf ||u < ǫ.

We now show that f is uniformly continuous. Select r > 0 such that ||Arf − f ||u < ǫ/3 and δ > 0such that |Arf(x)− Arf(y)| < ǫ/3 when |x− y| < δ. Then for any points such that |x− y| < δ,

|f(x)− f(y)| ≤ |f(x)−Arf(x)|+ |Arf(x)− Arf(y)|+ |Arf(y)− f(y)|

≤ ||f − Arf ||u + |Arf(x)− Arf(y)|+ ||Arf − f ||u

< ǫ/3 + ǫ/3 + ǫ/3 = ǫ.

By Theorem 3.18, Arf → f as r → 0 pointwise a.e. By unicity of the limit, we must have f = fa.e.

Problem 8.8Suppose that f ∈ LP (R). If there exists h ∈ Lp(R) such that

limy→0

||y−1(τ−yf − f)− h||p = 0,

we call h the (strong) Lp derivative of f . If f ∈ Lp(Rn), Lp partial derivatives of f are definedsimilarly. Suppose that p and q are conjugate exponents, f ∈ Lp, g ∈ Lq, and the Lp derivative ∂jfexists. Then ∂j(f ∗ g) exists (in the ordinary sense) and equals (∂jf) ∗ g.

Solution:First, we denote g(x) = g(−x). After a change of variables, we see that ||g||q = ||g||q. Denote the

Lp derivative of f as h ∈ Lp.

Then at any point x ∈ Rn, we have

3

|∂j(f ∗ g)(x)− (h ∗ g)(x)| = limt→0

∣

∣

∣

∣

∫

f(x+ tej − y)g(y)− f(x− y)g(y)

tdy −

∫

h(x− y)g(y)dy

∣

∣

∣

∣

≤ limt→0

∫

|g(y)| |t−1 (f(x+ tej − y)− f(x− y))− h(x− y)|dy

= limt→0

∫

|g(x− z)| |t−1 (f(z + tej)− f(z))− h(z)|dz

≤ limt→0

||τ−xg||q ||t−1(

τ−tejf − f)

− h||p

= ||g||q limt→0

||t−1(

τ−tejf − f)

− h||p = 0.

Problem 8.14 (Wirtinger’s Inequality)If f ∈ C1([a, b]) and f(a) = f(b) = 0, then

∫ b

a

|f(x)|2dx ≤

(

b− a

π

)2 ∫ b

a

|f ′(x)|2dx.

(By a change of variable it suffices to assume a = 0, b = 1/2. Extend f to [−1/2, 1/2] by settingf(−x) = −f(x), and then extend f to be periodic on R. Check that f , thus extended, is in C1(T) andapply Parseval identity.)

Solution:The first step is to show that, without loss of generality, we can assume that a = 0, b = 1/2.

Suppose the inequality holds for this specific case. Then via the change of variables x = 2(b−a)z+a,we obtain

∫ b

a

|f(x)|2dx = 2(b− a)

∫ 1/2

0

|f(2(b− a)z + a)|2dz

≤ 2(b− a)

(

1

2π

)2 ∫ 1/2

0

|d

dzf(2(b− a)z + a)|2dz

= (2(b− a))3(

1

2π

)2 ∫ 1/2

0

|f ′(2(b− a)z + a)|2dz

=

(

b− a

π

)2 ∫ b

a

|f ′(x)|2dx.

Next, we extend f to [−1/2, 1/2] by setting f(−x) = −f(x), and then extend f to be periodicon R in the obvious way. To check that f ∈ C1(T), it suffices to verify that f is differentiable andits derivative is continuous at 0; all other points of the form n/2, n ∈ Z are done similarly. By thedefinition of C1([0, 1/2]) given in Exercise 5.9, f has a one-sided derivative at the endpoint 0:

limt→0+

f(t)− f(0)

t= c.

4

The left-handed derivative can be computed using the symmetry of f :

limt→0−

f(t)− f(0)

t= lim

t→0+

f(−t)

−t= lim

t→0+

−(f(t)− f(0))

−t= c.

Hence f ′(0) exists and is equal to c. Next, we show that f ′(x) = f ′(−x). Indeed, d/dx(f(−x)) =−d/dx(f(x)) = −f ′(x), and on the other hand by the chain rule,

d

dx(f(−x)) = f ′(−x) · (−1).

Cancelling (−1) from both sides yields the result. We now show that the derivative is continuousat zero. Since f ∈ C1([0, 1/2]),

limt→0+

f ′(t) = c.

On the other hand,

limt→0−

f ′(t) = limt→0+

f ′(−t) = limt→0+

f ′(t) = c.

This completes the proof that f ∈ C1(T).

Now that we have reduced the problem to an easier case, we prove Wirtinger’s inequality. UsingParseval’s identity and integration by parts, we compute

∫ 1/2

−1/2

|f(x)|2dx = ||f ||2L2 =∑

k∈Z

∣

∣

∣

∫ 1/2

−1/2

f(x) e−i(2πkx)dx∣

∣

∣

2

=∣

∣

∣

∫ 1/2

−1/2

f(x)dx∣

∣

∣+

∑

k 6=0,k∈Z

∣

∣

∣

∫ 1/2

−1/2

f(x) e−i(2πkx)dx∣

∣

∣

2

=∑

k 6=0,k∈Z

∣

∣

∣

∫ 1/2

−1/2

f(x)−1

i(2πk)d (e−i(2πkx))

∣

∣

∣

2

=∑

k 6=0,k∈Z

∣

∣

∣

∫ 1/2

−1/2

f ′(x)

i(2πk)e−i(2πkx)dx

∣

∣

∣

2

=∑

k 6=0,k∈Z

1

4π2k2

∣

∣

∣

∫ 1/2

−1/2

f ′(x) e−i(2πkx)dx∣

∣

∣

2

≤1

4π2

∑

k 6=0,k∈Z

∣

∣

∣

∫ 1/2

−1/2

f ′(x) e−i(2πkx)dx∣

∣

∣

2

.

Since f = 0 at the endpoints, we see that∫ 1/2

−1/2f ′ = 0. Hence by Parseval’s identity we have

∫ 1/2

−1/2

|f(x)|2dx ≤1

4π2

∑

k∈Z

∣

∣

∣

∫ 1/2

−1/2

f ′(x) e−i(2πkx)dx∣

∣

∣

2

=1

4π2

∫ 1/2

−1/2

|f ′(x)|2dx.

5

Wirtinger’s inequality then follows by symmetry:

∫ 1/2

0

|f |2 =1

2

∫ 1/2

−1/2

|f |2 ≤1

2

(

1/2− 0

π

)2 ∫ 1/2

−1/2

|f ′|2 =

(

1/2− 0

π

)2 ∫ 1/2

0

|f ′|2.

Problem 8.16Let fk = χ[−1,1] ∗ χ[−k,k].

a. Compute fk(x) explicitly and show that ||f ||u = 2.b. f∨

k (x) = (πx)−2 sin 2πkx sin 2πx, and ||f∨k ||1 → ∞ as k → ∞. (Use Exercise 15a, and substitute

y = 2πkx in the integral defining ||f∨k ||1.)

c. F(L1) is a proper subset of C0. (Consider gk = f∨k and use the open mapping theorem.)

Solution:(a) By definition of convolution, we have

fk(x) =

∫ k

−k

χ[−1,1](x− y)dy.

For k > 0, k ∈ N, unravelling the definitions yields

fk(x) =

2 if −(k − 1) ≤ x ≤ k − 1,0 if x ≤ −(k + 1), or x ≥ k + 1,(1 + k)− x if k − 1 ≤ x ≤ k + 1,(1 + k) + x if −(k + 1) ≤ x ≤ −(k − 1).

We see that fk ∈ C0, and ||fk||u = 2.

(b) We use Exercise 15a to compute

(fk)∨(x) = (χ[−1,1] ∗ χ[−k,k])

∨ = (χ[−1,1])∨ (χ[−k,k])

∨ =1

(πx)2sin 2πkx sin 2πx.

By substituting y = 2πkx, we then obtain

||f∨k ||1 =

∫ ∞

−∞

∣

∣

∣

∣

1

(πx)2sin 2πkx sin 2πx

∣

∣

∣

∣

dx =4

π

∫ ∞

0

∣

∣

∣

∣

sin y

y

∣

∣

∣

∣

∣

∣

∣

∣

sin(y/k)

y/k

∣

∣

∣

∣

dy.

We know that for any y 6= 0,

limk→∞

∣

∣

∣

∣

sin y/k

y/k

∣

∣

∣

∣

= 1.

Also, for any B > 0, we have

∣

∣

∣

∣

sin y

y

sin y/k

y/k

∣

∣

∣

∣

χ[0,B] ≤ χ[0,B] ∈ L1(R).

6

By the Dominated Convergence Theorem, we obtain

limk→∞

4

π

∫ ∞

0

∣

∣

∣

∣

sin y

y

∣

∣

∣

∣

∣

∣

∣

∣

sin(y/k)

y/k

∣

∣

∣

∣

χ[0,B]dy =4

π

∫ ∞

0

∣

∣

∣

∣

sin y

y

∣

∣

∣

∣

χ[0,B]dy.

This leads to the conclusion that for all B > 0, we have

limk→∞

||f∨k ||1 ≥

4

π

∫ B

0

∣

∣

∣

∣

sin y

y

∣

∣

∣

∣

dy.

However, as shown in Exercise 2.59a (which was on last semester’s final exam), we know that∫∞

0|y−1 sin y|dy = ∞. Hence the right hand side can be made arbitrarily large, and ||f∨

k ||1 → ∞ ask → ∞.

(c) Suppose F : L1 → C0 is onto. By Corollary 8.27, we conclude that F : L1 → C0 is there-fore a bijection. By the Open Mapping Theorem, F−1 : C0 → L1 is a bounded linear operator. Inother words, there exists a C > 0 such that for all f ∈ C0, we have

||f∨||1 ≤ C||f ||u.

However, ||fk||u = 2 for all positive integers k, but ||f∨k ||1 can be made arbitrarily large as k → ∞.

This contradiction establishes that F(L1) is a proper subset of C0.

Problem 8.26The aim of this exercise is to show that the inverse Fourier transform of e−2π|ξ| on R

n is

φ(x) =Γ(1

2(n+ 1))

π(n+1)/2(1 + |x|2)−(n+1)/2.

a. If β ≥ 0, e−β = π−1∫∞

−∞(1 + t2)−1e−iβtdt. (Use (8.37).)

b. If β ≥ 0, e−β =∫∞

0(πs)−1/2e−se−β2/4sds. (Use (a), Proposition 8.24, and the formula

(1 + t2)−1 =∫∞

0e−(1+t2)sds.

c. Let β = 2π|ξ| where ξ ∈ Rn; then the formula in (b) expresses e−2π|ξ| as a superposition of

dilated Gauss kernels. Use Proposition 8.24 again to derive the asserted formula for φ.

Solution:(a) By (8.37), we have

e−2π|ξ| = F

(

1

π(1 + x2)

)

=

∫ ∞

−∞

π−1(1 + x2)−1e−i2πxξdx.

If we let ξ = (2π)−1β, we obtain

e−β = π−1

∫ ∞

−∞

(1 + t2)−1e−iβtdt.

7

(b) To justify using Fubini’s Theorem later on, we first of all compute:

∫ ∞

−∞

∫ ∞

0

|e−se−st2e−iβt|dsdt =

∫ ∞

−∞

∫ ∞

0

e−s(1+t2)dsdt =

∫ ∞

−∞

dt

1 + t2=

∣

∣

∣

∣

∞

−∞

arctan t = π < ∞.

We now compute from (a) and the formula (1 + t2)−1 =∫∞

0e−(1+t2)sds,

e−β =1

π

∫ ∞

−∞

(1 + t2)−1e−iβtdt =1

π

∫ ∞

−∞

∫ ∞

0

e−(1+t2)se−iβtdsdt.

By Fubini’s Theorem, we can swap the order of the integrals,

e−β =1

π

∫ ∞

0

∫ ∞

−∞

e−se−st2e−iβtdtds.

We now substitute z = β(2π)−1t and obtain

e−β =1

π

∫ ∞

0

e−s

(∫ ∞

−∞

2π

βe−π(4πs/β2)z2e−2πizdz

)

ds =2

β

∫ ∞

0

e−sF(e−π(4πs/β2)z2)(1) ds.

The Fourier transform of the Gaussian can be computed using Proposition 8.24, which yields

e−β =2

β

∫ ∞

0

e−s

(

β2

4πs

)1/2

e−β2/4sds =

∫ ∞

0

(πs)−1/2e−se−β2/4sds.

(c) As done in (b), before lauching into the proof, we compute an integral that will later allowus to use Fubini’s Theorem. The Gaussian is integrated using Proposition 2.53:

∫ ∞

0

∫ ∞

−∞

∣

∣

∣(πs)−1/2e−se−π2|ξ|2/se2πix·ξ

∣

∣

∣dξds =

∫ ∞

0

∫ ∞

−∞

(πs)−1/2e−se−π2|ξ|2/sdξds

=

∫ ∞

0

(πs)−1/2e−s

(

π

π2/s

)n/2

ds

=1

π(n+1)/2

∫ ∞

0

sn/2+1/2−1e−sds

=1

π(n+1)/2Γ(n/2 + 1/2) < ∞.

We now prove the claim. By part (b), letting β = 2π|ξ| we obtain

e−2π|ξ| =

∫ ∞

0

(πs)−1/2e−se−π2|ξ|2/sds.

8

We use this to write

(e−2π|ξ|)∨(x) =

∫ ∞

−∞

e−2π|ξ|e2πix·ξdξ =

∫ ∞

−∞

∫ ∞

0

(πs)−1/2e−se−π2|ξ|2/se2πix·ξdsdξ.

Applying Fubini, we exchange the order of integration

(e−2π|ξ|)∨(x) =

∫ ∞

0

(πs)−1/2e−s

∫ ∞

−∞

e−π2|ξ|2/se2πix·ξ dξds =

∫ ∞

0

(πs)−1/2e−s · (e−π2|ξ|2/s)∨(x) ds.

Proposition 8.24 gives us the inverse Fourier transform of a Gaussian, hence

(e−2π|ξ|)∨(x) =

∫ ∞

0

(πs)−1/2e−s( s

π

)n/2

e−s|x|2ds =1

π(n+1)/2

∫ ∞

0

s(n−1)/2e−s(1+|x|2)ds.

We substitute z = s(1 + |x|2) and obtain

(e−2π|ξ|)∨(x) =1

π(n+1)/2

(

1

1 + |x|2

)(n+1)/2 ∫ ∞

0

z1

2(n+1)−1e−zdz =

Γ(12(n+ 1))

π(n+1)/2(1 + |x|2)−(n+1)/2.

Problem 8.31Suppose a > 0. Use (8.37) to show that

∞∑

−∞

1

k2 + a2=

π

a

1 + e−2πa

1− e−2πa.

Then subtract a−2 from both sides and let a → 0 to show that∑∞

1 k−2 = π2/6.

Solution:Using the notation φ from (8.37), we have

∞∑

−∞

1

k2 + a2=

π

a2

∞∑

−∞

1

1 + (ka)2

=π

a2

∞∑

−∞

φ(k/a).

By the Poisson summation formula, we obtain

π

a2

∞∑

−∞

φ(k/a) =π

a2

∞∑

−∞

(φ(k/a))∧ .

Using Theorem 8.22b, we have

π

a2

∞∑

−∞

(φ(k/a))∧ =π

a2

∞∑

−∞

aφ(ka).

9

We now apply (8.37) to substitute for φ:

∞∑

−∞

1

k2 + a2=

π

a

∞∑

−∞

φ(ka)

=π

a

∞∑

−∞

e−2π|ka|

=π

a

(

1 + 2

∞∑

k=1

(

e−2πa)k

)

=π

a

(

1 + 2

(

1

1− e−2πa− 1

))

=π

a

1 + e−2πa

1− e−2πa.

Now that we have shown the desired identity, we subtract a−2 from both sides:

−1∑

−∞

1

k2 + a2+

∞∑

1

1

k2 + a2=

π

a

1 + e−2πa

1− e−2πa−

1

a2.

After simplification, we obtain

2∞∑

k=1

1

k2 + a2=

aπ − 1 + e−2πa(1 + aπ)

a2(1− e−2πa).

Letting a → 0, the left hand side goes to 2∑∞

1 k−2. For the right hand side, we repeatedly applyl’Hopital’s rule (three times in total):

lima→0

aπ − 1 + e−2πa(1 + aπ)

a2(1− e−2πa)= lim

a→0

π + πe−2πa − 2πe−2πa(1 + aπ)

2a(1− e−2πa) + 2πa2e−2πa

= lima→0

4aπ3e−2πa

2− 2e−2πa + 8aπe−2πa − 4π2a2e−2πa

= lima→0

4π3e−2πa − 8aπ4e−2πa

12πe−2πa − 24π2ae−2πa + 8π3a2e−2πa

=4π3

12π.

Therefore 2∑∞

1 k−2 = π2/3, which completes the proof.

Problem 8.35The purpose of this exercise is to show that the Fourier series of “most” continuous functions on T

do not converge pointwise.a. Define φm(f) = Smf(0). Then φ ∈ C(T)∗ and ||φ|| = ||Dm||1.

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b. The set of all f ∈ C(T) such that the sequence {Smf(0)} converges is meager in C(T). (UseExercise 34 and the uniform boundedness principle.)

c. There exists f ∈ C(T) (in fact, a residual set of such f ’s) such that {Smf(x)} diverges forevery x in a dense subset of T. (The result of (b) holds if the point 0 is replaced by any other pointin T. Apply Exercise 40 in §5.3.

Solution:

(a) We have

φm(f) = Smf(0) =

∫

T

f(y)Dm(y)dy.

By linearity of the integral, φm acts linearly on C(T), and

|φm(f)| ≤

∫

T

|f | |Dm|dy ≤ ||f ||u ||Dm||1.

Therefore, φm ∈ C(T)∗. To show ||φ|| = ||Dm||1, we apply the Riesz Representation Theoremfor Radon measures. Indeed, since Lebesgue measure is a Radon measure and Dm ∈ L1, we knowdµ = Dmdy is a Radon measure. Hence we can rewrite

φm(f) =

∫

T

fdµ.

By definition, d|µ| = |Dm|dy, hence by the Riesz Representation Theorem (7.17):

||φm|| = ||µ|| = |µ|(T) =

∫

T

|Dm|dy = ||Dm||1.

(b) Proceed by contradiction. Suppose the set of all f ∈ C(T) such that the sequence {φm(f)} con-verges is nonmeager in C(T). Then supm |φm(f)| < ∞ for all f in some nonmeager subset of C(T). Bythe uniform boundedness principle, supm ||φm|| < ∞. However, by Exercise 34, ||φm|| = ||Dm||1 → ∞as m → ∞. This is a contradiction, and hence the set of all f ∈ C(T) such that the sequence {φm(f)}converges is meager in C(T).

(c) Enumerate the rational {rk} inside (0, 1). This yields a countable dense subset of T ≃ R/Z.Define

Tjk(f) = Sjf(rk).

Then

Tjk(f) =

∫

T

f(y)Dj(rk − y)dy =

∫

T

f(y)τ−rkDj(−y)dy.

Denote Dj(−y) = Dj. We can repeat the same analysis done in (a) and obtain Tjk ∈ C(T)∗ and

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||Tjk|| = ||τ−rkDj||1 = ||Dj||1

.By the same argument of part (b), we conclude that for all rk, the set of all f ∈ C(T) such that

supj |Tjkf | < ∞ is meager in C(T).

We now apply the principle of condensation of singularities (Exercise 5.40). We have shown thatfor each k there exists a f ∈ C(T) (indeed, a residual set of f ’s) such that sup{|Tjkf | : j ∈ N} = ∞.By the principle of condensation of singularities, there is a residual set of f ’s in C(T) such thatsup{|Tjkf | : j ∈ N} = ∞ for all k. Hence we obtain we a residual set of f ∈ C(T) such that{Smf(rk)} diverges for each rk.

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