SISSA

Advanced Analysis - A Academic year 2019-2020

Proposed problems

1. Let X be a separable Banach space and Y a subspace of X. Show that Y , endowed with the induced norm, is separable.

2. Let X be a Banach space and Y a finite-dimensional subspace of X. Show that Y is closed.

3. Let (M, d) be a compact metric space. Show that M is complete and separable.

4. Let (M, d) be a complete metric space and {An, n ∈ N} a countable family of open and dense subsets of M . Show that the set

A .

= \

n∈N

An

is dense in M .

5. Let H be a real Hilbert space and a ∈ H a nonzero vector. Show that for every x ∈ H, we have

dist(x, {a}^{⊥}) =|(x, a)|

∥a∥ .

6. Consider the Hilbert space ℓ^{∞} with its usual norm ∥ · ∥_{ℓ}^{∞} and the sets c_{0} .

=
{(a_{n}) ∈ ℓ^{∞} : a_{n} → 0} and c .

= {(a_{n}) ∈ ℓ^{∞}: a_{n} → a ∈ R}. Show that c0 and c are
closed separable subspaces of ℓ^{∞}.

Answer. Let cc(N, R) be the elements with compact support (recall, we are dealing
with functions N → R). Then it is easy to see that c^{0}= cc(N, R) and that c^{c}(N, Q)
is a countable dense subspace of c_{c}(N, R), and so also of c0. Since c =S

q∈Q(c_{0}+ q)
we obtain that also c is separable.

7. Consider the Hilbert space ℓ^{2} and a real sequence (a_{n}) such that a_{n} > 0 for
every n ∈ N and an → +∞. Show that the set

A .

= (

u ∈ ℓ^{2}:X

n∈N

a_{n}|u_{n}|^{2}≤ 1
)

is a precompact subset of ℓ^{2}.

Answer. Let T : D(T ) → ℓ^{2} be the map x −→ y = ax where x, y, a : N → R and
D(T ) .

= (

u ∈ ℓ^{2}:X

n∈N

an|un|^{2}< ∞
)

.

Now, it is easy to see that the above map is invertible, and that ℓ^{2} ∋ y −→

x = T^{−1}y := ^{y}_{a} ∈ D(T ) ⊂ ℓ^{2} is in fact a compact operator. Then, since A =
T^{−1}D_{ℓ}2(0, 1) we conclude that A is relatively compact.

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8. Let H be a Hilbert space and C1, C2 two nonempty, closed and convex subsets
such that C1⊂ C2. Given x ∈ H, call PC_{i}x the projection of x on Ci and d(x, Ci)
the distance of x from Ci (i = 1, 2). Show that

∥PC_{1}x − PC_{2}x∥^{2}≤ 2 d(x, C1)^{2}− d(x, C2)^{2} , ∀x ∈ H.

9. Let H be a complex Hilbert space and T ∈ L(H) an operator such that ∥T ∥ ≤ 1.

Show that

(a) T x = x if and only if (T x, x) = ∥x∥^{2};
(b) ker(I − T ) = ker(I − T^{∗}).

Answer. In (a) the implication ⇒ is obvious. So let us consider ⇐ and let us
consider a nonzero x ∈ H s.t. (T x, x) = ∥x∥^{2}. Then ∥x∥^{2} ≤ ∥T x∥∥x∥ ≤ ∥x∥^{2},
where we used ∥T ∥ ≤ 1, implies ∥T x∥ = ∥x∥. Now, for bx = x/∥x∥ consider the
orthogonal decomposition

T x = (T x,x)b bx + (T x − (T x,bx)x) = x + (T x − (T x,b x)b bx) . Then, by ∥T x∥ = ∥x∥ and

∥T x∥^{2}= ∥x∥^{2}+ ∥T x − (T x,bx)bx∥^{2},
we conclude that T x − (T x,bx)x = 0, and so T x = x.b

10. Find a Banach space X and a subset S ⊆ X such that S is strongly closed but not weakly closed.

Answer. In infinite dimension, take S = {x ∈ X : ∥x∥ = 1}. In general, if
f ∈ C^{0}(X, R) is a continuous convex function with limx→∞f (x) = +∞, then for
Sr= {x ∈ X : f (x) = r} is strongly closed but , by repeating the proof in the special
case f (x) := ∥x∥, the weak closure is {x ∈ X : f (x) ≤ r}.

11. Find a Banach space X, a bounded closed subset S ⊆ X and a continuous function f : S → R such that

sup

x∈S

f (x) = +∞.

Answer. Recall that one of the exercises of Cuccagna’s notes states the following:

Let X be an infinite dimensional Banach space. Show that for any r ∈ (0, 1/2) there exists a sequence {vn} in X such that ∥vn∥X = 1 and the closed balls DX(vn, r) are pairwise disjoint. Show also that S∞

n=1DX(vn, r) is a closed set in X.

if we assume the above statement, let any g ∈ C^{0}(X, R) with g(0) ̸= 0 and
g(x) = 0 for ∥x∥ ≥ 1/2. Then consider

f (x) =

∞

X

n=1

ng x − vn

r

Then each term of the sum is zero outside DX(vn, r/2), so it is easy to conclude
that f ∈ C^{0}(X, R). Obviously we haveS∞

n=1D_{X}(v_{n}, r) ⊆ D_{X}(0, r + 1) =: S, so we
have the desired result.

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12. Let X be a Banach space and K ⊆ X a compact subset. Show that any sequence in K which converges weakly, actually converges strongly.

Answer. If false, there would be an example of xn⇀ x which does not converge
strongly to x. On the other hand, for any strongly convergence subsequence {xn_{k}},
we have necessarily xn_{k}

−−−−−→ x (because of xk→+∞ n_{k}⇀ x). But the fact that it is false
that xn

−−−−−→ x, implies that there exists an ϵn→+∞ 0> and a subsequence {xn_{k}} with

∥xn_{k}− x∥ ≥ ϵ0. By compactness, {xn_{k}} has a subsequence that converges strongly
at a point y ∈ K with ∥y − x∥ ≥ ϵ0. But we have just discussed the fact that we
must have y = x. So we get a contradiction.

13. Let (X, d) be a metric space. Given two subsets A, B ⊆ X, set dist(A, B) .

= inf{d(x, y) : x ∈ A, y ∈ B}.

a) Given x ∈ X and positive numbers 0 < ρ < r, show that there exists δ > 0 such that

dist(B(x, ρ), B(x, r)^{c}) ≥ δ.

b) Given a proper, nonempty, closed subset C ⊆ X, show that there exists a ball B(x, r) in X such that dist(B(x, r), C) > 0.

14. Let X, Y be Banach spaces and T ∈ L(X, Y ) a compact operator. Let (x_{n})
be a sequence in X weakly converging to x in X. Show that the sequence (T xn)
converges strongly to T x in Y .

Answer. We know that there is C > 0 such that ∥xn∥ < C for all n. Then (T xn)
is a sequence in K := T D_{X}(0, C), which is a compact subspace of Y . Since T is
continuous for the weak topologies, we know T x_{n} ⇀ T x. Then use the result in
Exercise 12.

15. Let α > 0 and consider the sequence of functions given by
u_{n}(x) .

= min{1, |x|^{−α}}χB(0,n)(x), n ∈ N, x ∈ R^{d}.

Study the convergence of (u_{n}) in the strong and weak (weak* if p = ∞) topology
of L^{p}(R^{d}) for p ∈ [1, ∞].

Answer. If we set u(x) := min{1, |x|^{−α}}, we have u ∈ L^{p}(R^{d}) for ap > d.

Now u(x) − u_{n}(x) = χ_{R\B(0,n)}(x)u(x) which converges monotonically to 0 for any
x ∈ R^{d}. For d/a < p < ∞, by dominated convergence, we obtain un

−−−−→ u inn→∞

L^{p}(R^{d}). If p = ∞, we have

0 ≤ u(x) − un(x) = χ_{R\B(0,n)}(x)u(x) ≤ |n|^{−α},
and so we get un

−−−−→ u in Ln→∞ ^{∞}(R^{d}). Obviously, the above implies also weak
convergence. For p ≤ d/a, it is easy to check that ∥un∥_{L}p(R^{d})

−−−−→ ∞, and so then→∞

sequence in not weakly convergent.

16. Let H be a Hilbert space, T ∈ L(H) and T^{∗} the adjoint of T .
(a) Show that ∥T^{∗}T ∥ = ∥T T^{∗}∥ = ∥T ∥^{2}.

(b) Show that T^{∗}T and T T^{∗} are selfadjoint operators.

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17. Let H be a Hilbert space and {Mk, k ∈ N} a countable collection of finite- dimensional subspaces of H. Call Pk the orthogonal projector on Mk (k ∈ N) and set

P .

=

∞

X

k=1

2^{−k}Pk.
Show that P is a compact operator in L(H).

Answer. First of all, we have ∥Pk∥ ≤ 1 for all k. Then

∥P − S_{N}∥ ≤

∞

X

k=N +1

2^{−k}= 2^{−N N →+∞}−−−−−→ 0, where S_{N} :=

N

X

k=1

2^{−k}P_{k}.
Since S_{N} is finite rank for any N , we get that P is compact.

18. Consider the sequence of functions given by un(x, y) =

cosx n

+ sinx n

(1+e^{−ny}^{2}), (x, y) ∈ I .

= [−1, 1]×[−1, 1], n ∈ N.

Study the convergence of (un) in the strong and weak topology of L^{p}(I) (weak* if
p = ∞).

Answer. First of all, we have un(x, y) − vn(x, y) =

cosx n

+ sinx n

e^{−ny}^{2} for
v_{n}(x, y) := cosx

n

+ sinx n

. It is straightforward that for p < ∞ we have

∥un− vn∥L^{p}(I) ≤ 2∥e^{−ny}^{2}∥L^{p}(0,1)≤ 2n^{−}^{2p}^{1}∥e^{−y}^{2}∥L^{p}(R)

−−−−→ 0.n→∞

Since

∥un− vn∥_{L}^{∞}_{(I)}≥ ∥vn(·, 0)∥_{L}^{∞}_{((0,1))}≥ |vn(0, 0)| = 1,
we obviously do not have un− vn

−−−−→ 0 strongly in Ln→∞ ^{∞}(I). However, for f ∈
L^{1}(I) we have

|⟨un− vn, f ⟩| ≤ 2 Z

I

e^{−ny}^{2}|f (x, y)|dxdy−−−−→ 0^{n→∞}

by dominated convergence, and so un−vn⇀ 0 weakly* in L^{∞}(I). Since (use bounds
of errors for alternating series)

|vn(x, y) − 1| ≤ cosx

n

− 1

+ | sinx n

| ≤ x^{2}
2n^{2} +x

n ≤ 3 2n we have vn

−−−−→ 1 in Cn→∞ ^{0}(I), and so in particular also for all L^{p}(I). So, summing
up, u_{n}−−−−→ 1 strongly in L^{n→∞} ^{p}(I) for p < ∞ and u_{n} ⇀ 1 weakly*, but not strongly,
if p = ∞.

19. Let H be a complex Hilber space, T ∈ L(H) and (xn) a sequence in H weakly
converging to x ∈ H. Show that the sequence (T x_{n}) converges weakly to T x.

Answer. We have

(T x_{n}, y) = (x_{n}, T^{∗}y)−−−−−→ (x, T^{n→+∞} ^{∗}y) = (T x, y) for all y ∈ H =⇒ T x_{n}⇀ T x.

4

20. Given x ∈ R, let B(x, 1) be the open unit ball of center x in R. Consider
a sequence (xn) in R and define the sequence of functions u^{n} .

= χB(x_{n},1), where
χ denotes the characteristic function. Study the strong and weak convergence
of the sequence (u_{n}) in the space L^{2}(R) (that is to say establish if the sequence
is converging in such topologies and, in affirmative case, find the limit), in the
following cases:

(a) xn→ 0;

(b) |xn| → +∞.

Answer. Notice that un(x) = χ_{B(0,1)}(x − xn) and that in Cuccagna’s notes it is
shown that if xn

−−−−→ ∞ then χn→∞ _{B(0,1)}(·−xn) ⇀ 0 in L^{2}(R). In the case x^{n} −−−−→ 0^{n→∞}

we have, instead, χ_{B(0,1)}(· − x_{n})−−−−→ χ^{n→∞} B(0,1) strongly in L^{2}(R), by the fact that
the group R^{d} is strongly continuous (but not continuous in the operator norm) in
L^{2}(R).

21. Let H be a Hilbert space endowed with the inner product ⟨·, ·⟩ and D a subset
of H such that lsp(D) is dense in H. Show that, given a bounded sequence (x_{n}) in
H, such that ⟨x_{n}, y⟩ → ⟨x, y⟩ for any y ∈ D, then x_{n} ⇀ x.

22. Let I = [0, 1] ⊆ R and consider the Hilbert space X = L^{2}(I, R). Set
(T u)(x) .

= Z x

0

u(t) dt.

Show that T ∈ L(X) and find the adjoint T^{∗} of T . Answer. Continuity is very
simple, since, in fact, T is a bounded operator from L^{2}(I, R) into C^{1/2}(I), by, for
x < y,

|T u(x) − T u(y)| ≤ Z y

x

|u(t)|dt ≤ |x − y|^{1}^{2}∥u∥_{L}2(I)

Notice that this implies that T D_{L}2(I)(0, 1) is relatively compact in C^{0}(I), and so
also in L^{2}(I), which implies that T is a compact operator. For the adjoint

Z 1 0

T u(x)v(x)dx = Z 1

0

dxv(x) Z x

0

u(t)dt = Z 1

0

dt u(t) Z 1

t

v(x)dx =⇒ T^{∗}v(x) =
Z 1

x

v(t)dt.

23. Consider the set E .

= {e^{n}, n ∈ N} in ℓ^{2} defined by
e^{n}(k) = δn,k.
Show that E is a Hilbert basis in ℓ^{2}.

24. Let U be a bounded family in L^{1}(R) and ρ ∈ Cc^{∞}(R). Show that the family
{ρ ⋆ u, u ∈ U } is equicontinuous.

Answer. Since ρ ∈ C_{c}^{∞}(R), ρ is uniformly continuous, and so

∀ ϵ > 0 ∃ δϵ> 0 s.t. |x1− x2| < δϵ=⇒ |ρ(x1) − ρ(x2)| < ϵ

Since there exists C > 0 s.t. ∥u∥L^{1}(R)< C for all u ∈ U , for |x1− x2| < δϵwe have

|ρ ⋆ u(x1) − ρ ⋆ (x2)| ≤ Z

|ρ(x1− y) − ρ(x2− y)| |u(y)|dy < Cϵ for all u ∈ U .

5

This implies that {ρ ⋆ u, u ∈ U } is equicontinuous.

25. Let H be a Hilbert space and T ∈ L(H). Show that T is compact if and only
if the adjoint T^{∗} is compact.

Answer. T is compact if and only if for any ϵ > 0 there exists a finite rank Tϵ

s.t. ∥T − Tϵ∥ < ϵ. Notice now, that ∥T^{∗}− T_{ϵ}^{∗}∥ = ∥T − Tϵ∥ < ϵ. So, if T_{ϵ}^{∗} is finite
rank, then we conclude the exercise. So let S be finite rank. Then

S =

n

X

j=1

(·, fj)gj.
In the special case S = (·, f_{1})g_{1}

(Su, v) = ((u, f1)g1, v) = (u, f1)(g1, v) = (u, (g1, v)f1) = (u, S^{∗}v).

So

S^{∗}v = (g1, v)f1= (v, g1)f1.
So, more generally

S^{∗}=

n

X

j=1

(·, g_{j})f_{j},
which implies that T_{ϵ}^{∗} is finite rank.

26. Let H be a Hilbert space on C, {ek, k ∈ N} an orthonormal system in H and
(λ_{k}) an element of ℓ^{1}(C). Set

T x .

=

∞

X

k=1

λk(x, ek)ek. Show that T is a compact operator in L(H).

27. Consider the Hilbert space E .

= L^{2}(R^{n}, C) and let K ∈ L^{2}(R^{n}× R^{n}, C). Define
(TKu)(x) .

= Z

R^{n}

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

Show that T_{K} ∈ L(E) and that TK is selfadjoint if and only if K(x, y) = K(y, x)
for any pair (x, y) ∈ R^{n}× R^{n}.

Answer. u, v ∈ C_{c}^{0}(R^{n}) we have
Z

R^{n}

(TKu)(x)v(x)dx = Z

R^{n}

Z

R^{n}

K(x, y)u(y) dyv(x)dx = Z

R^{n}

Z

R^{n}

K(x, y)v(x)dxu(y) dy

= Z

R^{n}

Z

R^{n}

K(y, x)v(y)dyu(x) dx = Z

R^{n}

u(x)TSv(x)dx

with S(x, y) = K(y, x). If the operator is selfadjoint, we conclude that Z

R^{n}

u(x)T_{S}v(x)dx =
Z

R^{n}×R^{n}

u(x)v(y)K(y, x)dxdy = Z

R^{n}

u(x)T_{K}v(x)dx

= Z

R^{n}×R^{n}

u(x)v(y)K(x, y)dxdy for all u, v ∈ C_{c}^{0}(R^{n}).

6

Since C_{c}^{0}(R^{n})N C_{c}^{0}(R^{n}) generates all L^{2}(R^{n}× R^{n}, C), from the above we conclude
Z

R^{n}×R^{n}

f (x, y)K(y, x)dxdy = Z

R^{n}×R^{n}

f (x, y)K(x, y)dxdy for all f ∈ L^{2}(R^{n}× R^{n}, C).

This implies K(x, y) = K(y, x) for a.a. (x, y) ∈ R^{n}× R^{n}.

28. Let H be a Hilbert space and (un) an orthonormal sequence in H. Show that (un) converges weakly to zero. Answer. For any f ∈ H we have

f =

∞

X

n=1

(f, u_{n})u_{n} with ∥f ∥^{2}=

∞

X

n=1

|(f, un)|^{2}.

Obviously the above implies (u_{n}, f )−−−−−→ 0 for any f ∈ H. This is equivalent to^{n→+∞}

un⇀ 0.

29. Let p ∈ [1, ∞[ and f ∈ L^{p}(R). Show that for every δ > 0 we have
meas ({x : |f (x)| > δ}) ≤ δ^{−p}∥f ∥^{p}_{p}.

Answer. This is the well known, and simple to prove, Chebyshev’s inequality.

30. Let E ⊆ R be a measurable set with finite measure, p ∈ ]1, ∞], (u^{n}) a sequence
in L^{p}(E) and u ∈ L^{p}(E) such that un ⇀ u for p < ∞ or⇀ if p = ∞. Prove that^{∗}
the sequence (un) is equiintegrable.

31. Let H be a real Hilbert space, M ⊆ X a closed subspace and P the orthogonal projector on M . Show that P is selfadjoint. Answer. Recall that P x ∈ M with

(P x − x, y) = 0 for all x ∈ H and y ∈ M .
Notice that H = M ⊕ M^{⊥}. Then, since

(x, (P^{∗}− 1)y) = 0 for all x ∈ H and y ∈ M =⇒ P^{∗}y = y for all y ∈ M .
So P^{∗}= P in M . Next,

0 = (P x, y) = (x, P^{∗}y) for all x ∈ H and y ∈ M^{⊥}.

So P^{∗}y = 0 for y ∈ M^{⊥}. Since we also have P y = 0 for y ∈ M^{⊥}, we conclude
P^{∗}= P everywhere in H.

32. Consider the space X = C0(R^{d}) .

= Cc(R^{d})^{∥·∥}^{∞}. Given S ∈ X^{′} define
U .

=O ⊆ R^{d} open: ⟨S, u⟩X^{′},X = 0 ∀u ∈ X with supp u ⊆ O .
Then introduce

N .

= [

O∈U

O (domain of nullity of S); supp S .

= R^{d}\ N (support of S).

(i) Given a ∈ R^{d}, set

Ta(u) = u(a).

Show that T_{a}∈ X^{′} and find its norm and support.

7

(ii) Let (an) be a sequence in R^{d}, consider the sequence (Sn) = (Ta_{n}) and the
series

S =

∞

X

n=1

3^{−n}Sn.

(a) Show that S ∈ X^{′} and find its norm and support.

(b) Show that there exists a subsequence (S_{n}_{k}) weakly* converging in X^{′}.

Answer. We have ∥Ta∥ = 1 as can be seen that more generally considering
σ_{N}f =

N

X

n=1

3^{−n}f (a_{n}) for σ_{N} :=

N

X

n=1

3^{−n}S_{n}

and, since we can always find f ∈ Cc(R^{d}) of norm 1 and with f (an) = 1 for all
n = 1, ..., N , we get

∥σN∥ =

N

X

n=1

3^{−n}= 3^{−1}1 − 3^{−N −1}

1 − 3^{−1} = 2^{−1}(1 − 3^{−N −1})
The implies that for all N < M we have

∥σ_{N}− σ_{M}∥ =

M

X

n=N +1

3^{−n}<

∞

X

n=N +1

3^{−n}= 3^{−N −1}3
2.

Hence, by completeness of X^{′}, there exists S limit of the above sum, with ∥σN∥−−−−−→^{N →+∞}

2^{−1} = ∥S∥. I skip the discussion of the support and go to the last question. Given
any sequence (a_{n}) in R^{d}there always exists a subsequence (a_{n}_{k}) which either a limit
a ∈ R^{d} or which goes to infinity. In the first case, for any f ∈ C_{0}(R^{d}), we have

Ta_{nk}f = f (an_{k})−−−−−→ f (a) = T^{k→+∞} af
while in the second case we have

T_{a}_{nk}f = f (a_{n}_{k})−−−−−→ 0,^{k→+∞}

since it is easy to see that C0(R^{d}) = {f ∈ C^{0}(R^{d}) : limx→∞f (x) = 0}. In the first
case Ta_{nk} ⇀ Ta while in the second Ta_{nk} ⇀ 0.

33. Consider the sequence of functions given by u_{n}(t) = sin(nt), with n ∈ N and
t ∈ I .

= [0, 2π]. Study the convergence of (u_{n}) in the uniform topology of C(I), in
the strong topology of L^{∞}(I) and in the weak* topology of L^{∞}(I) (that is to say
establish if the sequence is converging in such topologies and, in affirmative case,
find the limit). Answer. Clearly (sin(nt)) is not convergent strongly, while for any
f ∈ L^{1}(I) we have

Z

I

sin(nt)f (t)dt−−−−−→ 0^{n→+∞}

by the Riemann Lebesgue lemma, and so sin(nt) ⇀ 0 in the weak* topology of
L^{∞}(I).

8

34. Let X = C0(R^{2}, R), endowed with the uniform norm, and (a^{n}) a sequence in
R^{+}. Set

⟨fn, u⟩ .

= Z 2π

0

u(ancos θ, ansin θ) dθ, ∀n ∈ N, ∀u ∈ X.

Show that fn∈ X^{′} for every n ∈ N and find its norm and support.

Suppose an → 0+ and study the convergence of the sequence (fn) in the strong
and weak* topology of X^{′} (that is to say establish if the sequence converges in such
topologies and, in the affirmative case, find the limit).

Answer. We have f_{n} = 1
an

dx_{|∂D(0,a}_{n}_{)}, where dx_{|∂D(0,a}_{n}_{)} is the restriction of
the Lebesgue measure of R^{2} on the circle ∂D(0, a_{n}) and with ∥f_{n}∥(C_{0})^{′} = 2π. It is
elementary that for any u we have

Z 2π 0

u(ancos θ, ansin θ) dθ−−−−−→ u(0),^{n→+∞}

and so fn ⇀ δ(x) *–weakly . But it is easy to show that ∥fn− fm∥_{(C}_{0}_{)}′ ≥ 2π for
an̸= am.

35. Given α ∈ R and R > 0, consider the function u defined on R^{d} by
u(x) =

(|x|^{α}, x ̸= 0
0, x = 0.

Establish for which p ∈ [1, +∞] we have u ∈ L^{p}(B_{R}d(0, R)).

Answer. We need pα > −d. So, for example, if p = ∞, then α ≥ 0.

36. Let E ⊆ R^{d} be a measurable set, p, q ∈ [1, ∞[ and u ∈ L^{p}(E) ∩ L^{q}(E). Given
α ∈ [0, 1], set

1 r

=. 1 − α p +α

q.
Show that u ∈ L^{r}(E) and that

∥u∥_{L}r(E)≤ ∥u∥^{1−α}_{L}p(E)· ∥u∥^{α}_{L}q(E).

Answer. We have

∥u∥_{L}r(E)= ∥ |u| ∥_{L}r(E)= ∥ |u|^{1−α}|u|^{α}∥_{L}r(E)≤ ∥ |u|^{1−α} ∥

L

p

1−α(E)∥|u|^{α} ∥

Lα^{p}(E) by H¨older

= ∥u∥^{1−α}_{L}p(E)· ∥u∥^{α}_{L}q(E),
where we use the identity ∥|f |^{a}∥L^{q} = ∥f ∥^{a}_{L}aq.

37. Let I .

= [0, 1] and consider the sequence of functions given by
un(t) = e^{−nt}, t ∈ I, n ∈ N.

Study the convergence of the sequence (u_{n}) in the following spaces:

(i ) C^{0}(I) endowed with the uniform topology;

(ii ) L^{1}(I) endowed with the strong topology;

(iii ) L^{1}(I) endowed with the weak topology;

(iv ) L^{∞}(I) endowed with the strong topology;

9

(v ) L^{∞}(I) endowed with the weak* topology.

Answer. We have limn→+∞e^{−nt} = 0 if t > 0 and 1 if t = 0. So, clearly
the sequence is not convergent uniformly in [0, 1] (since the limit function is not
continuous). This also excludes convergence in L^{∞}(I) endowed with the strong
topology. We have

0 <

Z 1 0

e^{−nt}dt = n^{−1}
Z n

0

e^{−t}dt < n^{−1 n→+∞}−−−−−→ 0,

which implies that e^{−nt n→+∞}−−−−−→ 0 strongly, and so also weakly, in L^{1}(I). Finally,
if f ∈ L^{1}(I)

Z 1 0

e^{−nt}f (t)dt−−−−−→ 0,^{n→+∞}

by dominated convergence, and shows e^{−nt}⇀ 0 in L^{∞}(I) endowed with the weak*

topology.

38. Let E ⊆ R be a measurable set with finite measure and let m ∈ L^{∞}(E). Set
(T u)(x) .

= m(x) · u(x) for a.e. x ∈ E.

Given p, q ∈ [1, ∞[, with p ≥ q, show that T ∈ L(L^{p}(E), L^{q}(E)) and provide an
estimate of its norm.

Answer. We have

∥mu∥L^{q}(E)≤ ∥m∥L^{∞}(E)∥u∥L^{q}(E)= ∥m∥_{L}∞(E)∥1Eu∥_{L}q(E)≤ ∥m∥L^{∞}(E)|E|^{1}^{r}∥u∥L^{p}(E),
where |E| is the measure of E and ^{1}_{p} = ^{1}_{r}+^{1}_{q}.

39. Let X = Cc(R) and T : X → X be a linear application such that

∥T u∥_{L}1 ≤ ∥u∥_{L}1; ∥T u∥_{L}2 ≤ ∥u∥_{L}1 ∀u ∈ X.

Given r ∈ [1, 2], show that there exists ˜T ∈ L(L^{1}, L^{r}) such that ∥ ˜T ∥_{L(L}1,L^{r}) ≤ 1
and ˜T |X = T .

Answer. This is part of a more general result, called the Riestz interpolation Theorem. Here is just a consequence ot H¨older’s inequality, because for u ∈ Cc(R), by exercise 36, using 1

r = 1 −_{r}^{2}′

1 +

2
r^{′}

2

∥T u∥L^{r} ≤ ∥T u∥^{1−}

2 r′

L^{1} ∥T u∥

2 r′

L^{2}≤ ∥u∥^{1−}

2 r′

L^{1} ∥u∥

2 r′

L^{1} = ∥u∥_{L}1.

Then, by the density of C_{c}(R) in L^{1}(R), we get the unique extension ˜T ∈ L(L^{1}, L^{r})
with ∥ ˜T ∥_{L(L}1,L^{r})≤ 1 and ˜T |_{X} = T .

40. Consider the sequence of functions given by
un(x, y) = cos(nx)e^{−ny}, (x, y) ∈ I .

= [0, 2π] × [0, 2π], n ∈ N.

(a) Study the equicontinuity of (un) on I.

(b) Study the convergence of (u_{n}) in the uniform topology of C(I), in the strong
topology of L^{∞}(I) and in the weak* topology of L^{∞}(I).

10

Answer. It is not equicontinuous. Indeed, for y = 0 and for fixed m and n = 2mk we have

cos(0) − cos 2mk π

2m

= 1 − (−1)^{k}.

Since here _{2m}^{π} is arbitrarily close to 0 for m ≫ 1, we see that equicontinuity in
the origin is false. Since u_{n}(x, y)−−−−−→ 0 pointwise for y > 0 and u^{n→+∞} n(0, 0) ≡ 0,
there can be no uniform convergence in C(I), nor in the strong topology of L^{∞}(I)
(which would imply uniform convergence in C(I)). But there is weak* convergence
in L^{∞}(I) to 0, like in exercise 37.

41. Let X = C0(R^{2}, R) endowed with the uniform topology and consider the family
of subsets of R^{2}given by

Aα

=. (x, y) ∈ R^{2}: y > α|x|, x^{2}+ y^{2}< α^{−2} , α > 0.

Set

Tαu .

= Z

Aα

u(x, y) dxdy, α > 0.

(a) Show that Tα∈ X^{′} for every α > 0 and find its norm and support.

(b) Study the convergence of the family (Tα)α>0in the strong and weak* topol-
ogy of X^{′} when α → 0+ and when α → +∞.

Answer. We have

|T_{α}u| ≤
Z

Aα

|u(x, y)|dxdy ≤ Z

Aα

dxdy∥u∥_{L}^{∞} = |A_{α}|∥u∥_{L}^{∞}.

So ∥Tα∥ ≤ |Aα| and it is easy to show that there is equality. Finally,

|Aα| = 2π

2 − arctan α 1
α^{2}.
We obviously have Tα

−−−−−→ 0 strongly, while we cannot have weak* convergenceα→+∞

for α ↘ 0 because ∥Tα∥−−−−→ +∞.^{α→0}^{+}

42. Let I = [0, 1], X = C(I, R) and Y = L^{2}(I). Set
(T u)(x) .

= Z x

x^{2}

u(t) dt.

(a) Show that T ∈ L(X) and establish if T (B^{X}_{1} ) is relatively compact in X.

(b) Show that T ∈ L(Y ) and establish if T (B_{1}^{Y}) is relatively compact in Y .
Answer. One can write T = A − B with

Au(x) = Z x

0

u(t) dt

Bu(x) =
Z x^{2}

0

u(t) dt

11

Using H¨older inequality, for x1< x2,

|Au(x1) − Au(x2)| ≤
Z x_{2}

x_{1}

|u(t)|dt ≤ |x1− x2|^{1}^{2}∥u∥L^{2}(I)

|Bu(x1) − Bu(x2)| ≤
Z x^{2}_{2}

x^{2}_{1}

|u(t)|dt ≤ |x^{2}_{1}− x^{2}_{2}|^{1}^{2}∥u∥L^{2}(I) ≤√

2|x1− x2|^{1}^{2}∥u∥L^{2}(I).
Then T : Y → C^{1}^{2}(I) is bounded, and hence T (B_{1}^{Y}) is bounded in X and equicon-
tinuous. This means that T : Y → X is compact and, a fortiori, also the other
maps T : Y → Y and T : X → X.

43. Consider the sequence of functions given by un(x, y) = sin

n^{2}x
n + 1

e^{y/n}, (x, y) ∈ I .

= [0, 2π] × [0, 2π], n ∈ N.

(a) Study the equicontinuity of (un) on I.

(b) Study the convergence of (un) in the uniform topology of C(I); in the strong
and in the weak* topology of L^{∞}(I).

Answer. Using

1

1 +_{n}^{1} = 1 − 1

n+ O 1
n^{2}

we have

u_{n}(π/2, 0) = sin

n

π 2

1 +_{n}^{1}

= sin

nπ

2 −π

2 + O 1 n

= − cos

nπ

2 + O 1 n

sinπ 2

= − cos

nπ

2 + O 1 n

= − cos nπ

2

cos

O 1

n

+ sin nπ

2

O 1

n

= − cos nπ

2

+ O 1 n

and

u_{n}(π/2 − π/(2n), 0) = sin n

π
2−_{2n}^{π}

1 +_{n}^{1}

!

= sin

nπ

2 − π + O 1 n

= − sin

nπ

2 + O 1 n

. For n = 2k we have

un(π/2 − π/(2n), 0) − un(π/2, 0) = cos (kπ) − sin

kπ + O 1 k

+ O 1 k

= cos (kπ) − sin (kπ) + O 1 k

= (−1)^{k}+ O 1
k

.

This excludes equicontinuity in (π/2, 0), which would require that for any ϵ > 0 there exists δ > 0 such that

x − π

2

< δ =⇒ |u_{n}(x, 0) − u_{n}(π/2, 0)| < ϵ for all n.

12

There is no strong convergence since otherwise we would have equicontinuity, which
we have just excluded. On the other hand, for any f ∈ C^{0}(I) we have

Z

I

un(x, y)f (x, y)dxdy = Z 2π

0

dye^{y/n}
Z 2π

0

sin

n^{2}x
n + 1

f (x, y)dx

= Z 2π

0

dye^{y/n}
Z 2π

0

sin

nx

1 − 1

n+ O 1
n^{2}

f (x, y)dx

= Z 2π

0

dye^{y/n}
Z 2π

0

sin

nx − x + O 1 n

f (x, y)dx

= Z 2π

0

dye^{y/n}
Z 2π

0

sin (nx) cos

x + O 1 n

f (x, y)dx

− Z 2π

0

dye^{y/n}
Z 2π

0

cos (nx) sin

x + O 1 n

f (x, y)dx =: I_{n}+ II_{n}.
We claim that I_{n} −−−−−→ 0 and II^{n→+∞} _{n} −−−−−→ 0. We will prove only the first limit,^{n→+∞}

since the second is similar. We have In=

Z 2π 0

dye^{y/n}
Z 2π

0

sin (nx) cos (x) f (x, y)dx + O 1 n

. Then, for f (x, y) = a(x)b(y)

In≤ Z 2π

0

dye^{2π}|b(y)|

Z 2π 0

sin (nx) cos (x) a(x)dx

+ O 1 n

. Since by Riemann-Lebesgue

Z 2π 0

sin (nx) cos (x) a(x)dx−−−−−→ 0^{n→+∞}

we get In

−−−−−→ 0 by dominated convergence. This extends by linearity for alln→+∞

f ∈ L^{1}(0, 2π)N L^{1}(0, 2π) and, since the un are uniformly bounded in n, by density,
to all f ∈ L^{1}(I).

44. Let X = C0(R^{2}, R) endowed with the uniform norm and consider the family
of subsets of R^{2}given by

Aα

=. (x, y) ∈ R^{2}: x > 0, y > α|x|, x^{2}+ y^{2}< α^{2} , α > 0.

Set

Tαu .

= 1
α^{2}

Z

A_{α}

u(x, y) dxdy, α > 0.

(a) Show that T_{α}∈ X^{′} for any α > 0 and find its norm and support.

(b) Establish if the family (T_{α})_{α>0}converges in the strong and weak* topology
of X^{′} when α → 0+ and, in affirmative case, determine the limit T0.
(c) Find norm and support of T0.

45. Let I = [0, 1], X = C(I, R) and α(x) .

= min{1, 2x}. Set (T u)(x) .

= Z α(x)

0

|u(t)|^{2}dt.

Establish if T ∈ L(X) and if T (B_{1}^{X}) is relatively compact in X.

13

46. Consider the following family of Cauchy problems:

(y^{′}= _{1+ty}^{1} t > 0
y(0) = 1 + ^{1}_{n} n ∈ N.

(a) Show that for every n ∈ N there exists a solution y^{n}(·) defined on the whole
R^{+}.

(b) Show that the sequence (yn) amdits a subsequence uniformly converging
on each compact subinterval of R^{+}.

Answer. First of all, for y(0) > 0 the corresponding solution is positive and strictly growing. Suppose that one of these solutions, with initial value y0> 0, has maxi- mum forward time of existence [0, t1). We need to show that t0= +∞. If not and t1 < +∞, then, by monotonicity, the limit lim

t→t^{−}_{1}

y(t) exists. If this limit is finite and equals to y1, by 0 < y0< y1 we have y1∈ R+. but then, considering

(y^{′}=_{1+ty}^{1} t > 0
y(t1) = y1

it is easy to see that we can extend the previous equation beyond [0, t_{1}). So we get
a contradiction, and lim

t→t^{−}_{1}

y(t) = +∞. Let us show that also this is impossible. Let 0 < t2< t1 with y(t2) ≥ 1. Then, for t2< t < t1 we have y(t) > y(t2)

y(t) − y(t2) = Z t

t2

y^{′}(s)ds =
Z t

t2

1

1 + sy(s)ds ≤ Z t

t2

1

1 + t_{2}dt = t − t2

1 + t_{2}

t→t^{−}_{1}

−−−→ t1− t2

1 + t_{2}.
Hence +∞ = lim

t→t^{−}_{1}

(y(t)−y(t2)) ≤ t1− t2

1 + t2

, which obviously is a contradiction. So we
conclude that if y(0) = y_{0}> 0, the corresponding solution is defined in [0, +∞). The
uniform convergence on compact intervals, is a consequence of the ”well posedness”

of solutions of the Cauchy problem for ODE’s, which is a very general fact. Suppose
that (y0n) is a sequence in R^{+} with y0n

−−−−−→ yn→+∞ 0 with y0 ∈ R+. Then we can prove that the yn

−−−−−→ y in Cn→+∞ ^{0}([0, T ]) for any T > 0, where yn(t) = y0n and
y(0) = y0 using the Growall inequality. From

y^{′}− y_{n}^{′} = f (t, y) − f (t, y_{n})
we get, after some elementary computations,

|y(t) − yn(t)| ≤

1 +

Z t 0

An(s)ds

|y0− y0n| + Z t

0

An(s)|y(s) − yn(s)|ds with An(s) :=|f (s, y(s)) − f (s, yn(s))|

|y(s) − yn(s)| . Notice now that

An(s) ≤ Z 1

0

|∂yf (s, yn(s) + τ (y(s) − yn(s)))|dτ ≤ sup {|∂yf (s, y)| : s ∈ [0, T ] and y ≥ 0} ≤ T So we get

|y(t) − yn(t)| ≤ 1 + T^{2} |y0− y0n| + T
Z t

0

|y(s) − yn(s)|ds.

14

Now, Gronwall’s inequality yields

|y(t) − y_{n}(t)| ≤ e^{T t} 1 + T^{2} |y0− y_{0n}|
which yields yn

−−−−−→ y in Cn→+∞ ^{0}([0, T ]).

47. Consider the sequence of functions given by un(x, y) = sin

nx n + 1

(1 + e^{−n|y|}), (x, y) ∈ I .

= [ − 1, 1] × [ − 1, 1], n ∈ N.

(a) Study the equicontinuity of (u_{n}) on I.

(b) Study the convergence of (un) in the uniform topology of C(I), in the strong
topology of L^{∞}(I) and in the weak* topology of L^{∞}(I).

48. Let I = [0, 1], X = C^{0}(I) and m ∈ X. Set
(Tmu)(x) .

= m(x)u(x), u ∈ X, x ∈ I.

Show that Tm ∈ L(X) and that it is compact if and only if m(x) = 0 for every x ∈ I.

Answer. We have σ(Tm) = m(I). We must have 0 ∈ m(I) and m(I)\{0} must be discrete and m(I) must be connected. So, summing up, m(I) = {0}, which implies m ≡ 0 and Tm= 0.

49. Let B the closed unit ball in R, endowed with the euclidean norm ∥ · ∥. Define un(x) .

= |sin(∥x∥)|^{n}^{1} n ∈ N.

Study the equicontinuity of the familyun, n ∈ N on B. Answer. Recall that the
condition of the Ascoli Arzela Theorem are both sufficient and necessary. Obviously
the above sequence is is bounded in C^{0}(B). Notice that pointwise we have

|sin(|x|)|^{1}^{n} −−−−−→^{n→+∞} 1 if x ̸= 0
0 if x = 0

and this excludes the existence of a subsequence converging uniformly. So the se- quence is not equicontinuous.

50. Consider the sequence of functions given by
un(x, y) = e^{−}^{n+1}^{ny}

(1 + e^{−nx}^{2}), (x, y) ∈ I .

= [ − 1, 1] × [ − 1, 1], n ∈ N.

Study the convergence of (u_{n}) in the uniform topology of C(I), in the strong topol-
ogy and in the weak* topology of L^{∞}(I).

51. Let φ ∈ C_{c}(R) and (an) a sequence in R. Define
un(x) .

= φ(x − an), x ∈ R, n ∈ N.

a) Show that u_{n} ∈ L^{p}(R) for every p ∈ [1, ∞].

b) Study the relative compactness of the sequence (u_{n}) in the strong and in the
weak topology of L^{p} (weak* if p = ∞). That is to say: establish if and for which
p ∈ [1, ∞] there exists a converging subsequence in such topologies.

52. Let I = [0, 1] ⊂ R and B .

=u ∈ C^{1}(I) : ∥u^{′}∥L^{2}(I) ≤ 1 .

15

a) Show that B is an equicontinuous family.

b) Given a sequence (un) in {u ∈ B : u(0) = 0, u(1) = 1}, show that there exist
u ∈ C^{0}(I) and a subsequence (un_{k}) which converges uniformly to u.

c) Show by a counterexample that property b) does not hold in B.

53. Let B the closed unit ball in R^{d}, endowed with the euclidean norm ∥ · ∥. Set
un(x) .

= e^{−n∥x∥} n ∈ N.

Study the equicontinuity of the familyun, n ∈ N on B.

54. Consider the sequence of functions given by un(x, y) = minn

n, |x|^{−}^{1}^{2}o

sin ny n + 1

, (x, y) ∈ I .

= [ − 1, 1] × [ − 1, 1], n ∈ N.

Study the convergence of (u_{n}) in the strong and weak topology (weak* if p = ∞)
of L^{p}(I).

55. Let φ ∈ C_{c}(R) with supp φ ⊆ [−1, 1], φ ≥ 0 e R

Rφ dt = 1. Consider the Dirac sequence given by

ρ_{n}(t) .

= nφ(nt) ∀t ∈ R ∀n ∈ N and let (an) be a sequence in R. Set

u_{n}(t) .

= ρ_{n}(x − a_{n}) ∀t ∈ R ∀n ∈ N.

a) Show that un ∈ L^{p}(R) for every p ∈ [1, ∞] and for every n ∈ N.

b) Considering the cases an = n and an = n^{−2}, study the convergence of the
sequence (un) in the strong and weak topology of L^{p} (weak* if p = ∞).

56. Let I = [0, 1] ⊂ R, X = C^{0}(I) and Y = L^{1}(I). Set
T u(x) .

= Z x

0

xyu(y) dy.

a) Show that T ∈ L(X) and T ∈ L(Y ).

b) Establish if T is compact in L(X) and in L(Y ), explaining the reasons.

57. Let (ρn)_{n∈N}be a regularizing sequence in R. Study the equiintegrability of the
following families:

a) fn = ρn, n ∈ N;

b) gn = ρ^{′}_{n} , n ∈ N;

c) hn

= ρ. 1⋆ ρn, n ∈ N.

58. Let α > 0 and consider the sequence of functions given by
u_{n}(x) .

= min{1, |x|^{−α}}χ_{B(0,n)}(x), n ∈ N, x ∈ R^{d}.

Study the strong and weak (weak* if p = ∞) convergence of (un) in the spaces
L^{p}(R^{d}) for p ∈ [1, ∞].

59. Let Q .

= [−1, 1]^{3}⊆ R^{3} and set
f (x1, x2, x3) =

( x1x^{2}_{2}x^{3}_{3}−1

, x1x2x3̸= 0
0, x_{1}x_{2}x_{3}= 0.

- Establish for which p ∈ [1, ∞] we have f ∈ L^{p}(R^{3});

- establish for which p ∈ [1, ∞] we have f ∈ L^{p}(Q);

16

- establish for which p ∈ [1, ∞] we have f ∈ L^{p}(R^{3}\ Q).

60. Let E ⊆ R be a measurable set, pi∈ [1, ∞], fi ∈ L^{p}^{i}(E) for i = 1, . . . , n, and
r ∈ [1, ∞] given by

1 r

=.

n

X

i=1

1 pi

. Show that

n

Y

i=1

fi∈ L^{r}(E)
and that the following inequality holds:

n

Y

i=1

fi

L^{r}(E)

≤

n

Y

i=1

∥fi∥_{L}pi(E).

61. Let (ρn)_{n∈N} be a regularizing family in R and f ∈ C^{0}(R). Set
f_{n}(x) .

= (ρ_{n}⋆ f )(x), x ∈ R.

Show that the definition is well posed and that the sequence (fn) converges uni- formly to f on any compact subset K ⊆ R.

62. Let I = [−1, 1] ⊆ R and (un) a sequence in C^{2}(R) such that
(a) u_{n} is convex on R for every n ∈ N;

(b) There exists K ≥ 0 such that |u_{n}(0)| + |u^{′}_{n}(t)| ≤ K for every t ∈ I and
for every n ∈ N.

(1) Show that the sequence (u^{′}_{n}) is relatively compact in L^{1}(I).

(2) Show that there exists a subsequence (un_{k}) and a map u ∈ C^{0}(I) such that
(un_{k}) converges uniformly to u on I.

63. Let I = [0, 1] ⊆ R and {e^{n}, n ∈ N} a Hilber basis L^{2}(I). Set
(em⊗ en)(x, y) .

= em(x)en(y); m, n ∈ N, (x, y) ∈ I × I.

Show that the family {e_{m}⊗ en; m, n ∈ N} is a Hilbert basis in L^{2}(I × I).

Answer. It is an orthonormal family. Now, for u, v ∈ L^{2}(I), we have
X

n,m∈N

|(u ⊗ v, e_{m}⊗ e_{n})|^{2}= X

n,m∈N

|(u, e_{m})|^{2}|(v, e_{n})|^{2}= X

m∈N

|(u, e_{m})|^{2}X

n∈N

|(v, e_{n})|^{2}

= ∥u∥^{2}_{L}2(I)∥v∥^{2}_{L}2(I) = ∥u ⊗ v∥^{2}_{L}2(I×I)

This means that each u ⊗ v is in the closed space spanned by the above orthonormal
family, hence also the linear conmbinations of these elements. Since the latter are
dense in L^{2}(I × I), it follows that the space generated by the orthonormal family is
L^{2}(I × I), and thus the orthonormal family is a Hilbert basis.

64. Let I .

= [ − 1, 1] ⊆ R and consider the sequence of functions given by:

un(t) = e^{−n}· e^{nt}^{2}; t ∈ I, n ∈ N.

Study the convergence of the sequence (u_{n}) in the following spaces:

(i ) C^{0}(I) with uniform topology;

17

(ii ) L^{1}(I) with strong topology;

(iii ) L^{1}(I) with weak topology;

(iv ) L^{∞}(I) with strong topology;

(v ) L^{∞}(I) with weak* topology.

65. For every n ∈ N set fn(x) = sinx

n

; gn(x) = sin n^{2}x ; hn(x) = sin

nx n + 1

; x ∈ [0, 2π].

Study the equicontinuity of the sequences {f_{n}, n ∈ N}, {gn, n ∈ N} e {hn, n ∈ N}

on [0, 2π].

66. Let I = [0, 1] ⊂ R and, for every n ∈ N, consider the subintervals of the form
I_{n}^{m} .

= m n,m + 1

n

, m = 0, 1, . . . , n − 1.

Then set

un(t) .

= (−1)^{m}for t ∈ I_{n}^{m}.

Study the strong and weak convergence of the sequence (un) in L^{2}(I).

67. Let D be te unit disk in C. Study the equicontinuity of the following families of functions in C(D):

(i) {fa(z) = e^{iaz}, a ∈ R};

(ii) {fa(z) = e^{i}^{z}^{a}, a ∈ R a ̸= 0};

(iii) {fa(z) = e^{iaz}, a ∈ R, |a| > 1};

(iv) {fa(z) = e^{iaz}, a ∈ R, |a| < 1}.

68. Let X = C([0, 1], R) and (a^{n}) a sequence in [0, 1]. Set

⟨f_{n}, u⟩ .

= u(a_{n}), ∀n ∈ N, ∀u ∈ X.

Show that f_{n}∈ X^{′}for every n ∈ N and that there exists a subsequence (fnk) which
converges in the topology σ(X^{′}, X).

69. Study the equicontinuity of the following families in C(I) (I ⊆ R)).

(i) {fa(x) = e^{ax}, a ∈ R}, I = R;

(ii) {fa(x) = a(1 − x)^{2}, a ∈ R^{+}}, I = [−1, 1];

(iii) {f_{a}(x) = x^{−a}, a ∈ R^{+}, }, I = ]1, +∞[;

(iv) {f_{a}(x) = x^{−a}, a ∈ R^{+}, }, I = ]0, +∞[.

70. Let p ∈ [1, ∞[. Consider the space X = L^{p}([0, 1]) and set
(T u) (x) =

Z x 0

u(t) dt.

(i) Show that T ∈ L(X) and that ∥T ∥_{L(X)}≤
p^{1}^{p}−1

.

(ii) Given a sequence (u_{n}) in X weakly converging to u in X, show that the
sequence (T u_{n}) converges strongly to T u in X.

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Answer. The bound follows from

|T u(x)| ≤ Z x

0

|u(t)|dt ≤ x^{p′}^{1}∥u∥_{L}p([0,1])

and from

∥x^{p′}^{1}∥L^{p}([0,1])=

Z 1 0

x^{p′}^{p}dx

^{1}p

= 1

p
p^{′} + 1dx

!^{1}_{p}

=

p^{1}^{p}−1 1

1
p^{′} +^{1}_{p}dx

!^{1}_{p}

=
p^{1}^{p}−1

.
For part (ii), the result follows from the fact that T : L^{p}([0, 1]) → L^{p}([0, 1]) is
compact. The case p = 1 is discussed in Cuccagna’s notes. The case p > 1 is easier
because we have for any x_{1}< x_{2}

|T u(x1) − T u(x2)| ≤
Z x_{2}

x_{1}

|u(x)|dx ≤ ^{p′}p|x1− x2|∥u∥L^{p}([0,1]).

From this and Ascoli Arzela we conclude that T : L^{p}([0, 1]) → C^{0}([0, 1]) is compact
for p > 1 and so, at fortiori, also T : L^{p}([0, 1]) → L^{p}([0, 1]) is compact.

71. Let C > 0, p ∈ [1, ∞[, α ∈ ]0, 1[ and B .

= {x ∈ R^{d}: ∥x∥ ≤ 1}. Consider the set
U .

= {u ∈ C(B) : u(0) = 0, |u(x) − u(y)| ≤ C|x − y|^{α} ∀x, y ∈ B} .
Show that U is relatively compact in L^{p}(B).

Answer. It is immediate that U is bounded and equicontinuous and so relatively
compact in C^{0}(B), and hence also in L^{p}(B).

72. Let I .

= [0, 1] and (u_{n}) a sequence in C^{1}([0, 1]) such that

|un(0)| + Z

I

|u^{′}_{n}(t)| dt ≤ 1 ∀n ∈ N.

Show that there exist a subsequence (un_{k}) and a map u ∈ L^{1}(I) such that un_{k}→ u
strongly in L^{1}(I).

73. Let E ⊆ R^{d} be ameasurable set such that 0 < m(E) < +∞. For every
p ∈ [1, +∞[ and for every f ∈ L^{p}(E) set

Np[f ] .

=

1

m(E) Z

E

|f (x)|^{p}

^{1}_{p}
.

Show that Np[·] is a norm on L^{p}(E) and that, if 1 ≤ p ≤ q < +∞, we have
Np[f ] ≤ Nq[f ] ∀f ∈ L^{q}(E).

74. Let X be a Banach space and set K(X) .

= {T ∈ L(X) : T is compact}. Show that K(X) is closed in L(X).

75. Let X = C0(R^{2}) and (an) a sequence in R^{+}. For every n ∈ N and for every
u ∈ X set

Tn(u) =
Z +a_{n}

−an

u(x, nx) dx.

Show that Tn ∈ X^{′} for every n ∈ N a find its norm and support. Study the
convergence of the sequence (Tn) in the strong and weak* topology of X^{′} in the
cases a_{n} = 1 + n^{2} and a_{n}= e^{−}^{n}^{1}.

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