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MMATH96037/MATH97062/MATH97173 Functional Analysis (2021)
发布时间:2022-05-07
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MMATH96037/MATH97062/MATH97173 Functional Analysis (2021)
1. (a) Prove that the space of polynomials on the unit interval [0, 1] with the supremum norm is a linear metric space. (6 marks)
(b) Prove or disprove that the space in (a) is :
(i) separable; (5 marks) (ii) complete. (4 marks)
(c) Prove or disprove that the space l4 is a Hilbert space. (5 marks) (Total: 20 marks)
2. (a) Let X be a proper closed subspace of the space lp , p e (1, o). For p˜ > p, consider the norm of lp˜ and of lp restricted to X and assume they are well defined on X. Prove that, if the space X is finite dimensional, then these norms are equivalent. By example show that if the space X is infinite dimensional, then the two norms on X are not necessarily equivalent.
(7 marks)
(b) Prove that a unit ball in the linear metric space dual to the space ((([0, 1]), |.|u ) of continuous functions with supremum norm | . |u , is infinite dimensional and not separable.
(6 marks)
(c) Consider the following linear space
W = {(xj e R)j∈N : x3 = 0;
Ixj I2 < o}
j∈N
with a norm
|x| = │ j
1
Ixj I2 ← ≠ .
3. (a) Let Hn (x), n e N, be Hermite polynomials associated to the measure
dµ = e一
x≠ dλ.
For what values of ξ e R the following operator is contractive in L2 (R, µ)
Tn f (x) = Hk (x) + Kξ f (x) = Hk (x) + ξHn (x) Hn (y)f (y)µ(dy) (*)
with k n. (5 marks) (b) Prove that for ξ = 1 the operator in (*) is not strictly contractive but nevertheless has a fixed point. (5 marks)
(c) Let (pj )j∈N be a sequence of distinct prime numbers. Prove that the operator defined as
follows
Kf = 1 Hn (x) Hn (y)f (y)µ(dy)
is compact in L2 (R, µ) . (10 marks)
Hint: The k-th prime number asymptotically is of order
pk ~ k log(k).
(Total: 20 marks)
4. (a) Prove that the following subspace
V = {f e (([0, 1]) : cc e R Vx e [1/3, 2/3] f (x) = c}
is a closed linear subspace in the space of continuous functions (([0, 1]) with supremum norm | . |u . (5 marks) (b) Prove or disprove that V is nowhere dense in ((([0, 1]), | . |u ). (6 marks)
(c) For p e (1, o) and a sequence ρ = (ρj e (0, o))j∈N satisfying ↘j∈N ρj = 1, let
lp (ρ) = {(xn e R)n∈N : Ixj Ip ρj < o}.
j
Prove that, if for some (zj e R)j∈N we have
Vx e lp (ρ)
zj xj ρj < o,
j
then there exists uncountably many q e (1, o) such that
Izj Iq ρj < o.
j
(9 marks) (Total: 20 marks)
5. (a) Explain what does it mean that a Banach space (X, |.|X ) is compactly embeded into another Banach space (Y, | . |Y )
(4 marks)
(b) Define Wk,p (Ω) space and show that it is a separable Banach space.
(6 marks)
(c) Let Ω = [0, 1]d c Rd , d > 3, and for any x e Ω define
Tf (x) = e一 ←从 ì)≠ f (y)λ(dy)
Ω
for any f : Ω - R for which the right hand side is well defined. Using Sobolev inequality or otherwise, show that the following set
{Tf : |f |2 < 2}
is compact in L2 (Ω).
(10 marks) (Total: 20 marks)
MMATH96037/MATH97062/MATH97173
Functional Analysis (Solutions)
l. (a) In the space of polynomials we are given natural addition of vectors and multiplication by a number as for functions as follows. For
P(j)(t) =
α)tk
k>〇
with a convention that 3n ∈ N Vk > n α) = 0, we have
╱P(1) + P(2)← (t) = (α + α
)tk
k
and
╱ λ . P(1)← (t) = (λα )tk
k
One needs to show that this operations are continuous. Since the metric
ρ(v, v/) = lv e v/l
is given by the norm, the continuity is immediate by the following general argument lv e w e v/ e w/l < lv e v/l + lw e w/l
for any vectors v, v/, w, w/, using commutativity of addition of vectors and Minkowski inequality. Similarly for multiplication of a vector by a number, we have
lλ o v e λ/ o v/l = l(λ _ λ/) o v e λ/ o (v e v/)l < Iλ _ λ/I . lvl + Iλ/I . lv e v/l
< max{lvl, Iλ/I} ╱Iλ _ λ/I + lv e v/l、
(b)
(i) The space is separable. This is because the countable set of monomials Xn = tn , n ∈ Z+ is the Hamel basis of the space, (by definition of polynomials), and we can approximate coefficients by rational numbers, that is for any ε > 0 and any polynomial P (t) = k=〇,..,n αktk and rationals qk ∈ Q, k = 0, .., n, we have
l
αktk _
qktk l <
Iαk _ qkI
k=〇,..,n k=〇,..,n k=〇,..,n
and choosing maxk=〇,..,nIαk _ qkI < ε we get
lP (t) _ qktk l < ε.
k=〇,..,n
(ii) The space is not complete. For example a sequence
n tn
2n
k=〇
is Chauchy with respect to the sup norm on [0, 1] and the sequence converges in the sup norm to a bounded continuous function (1 _
)-1 .
(c) Since the norm of a Hilbert space by definition is given by a scalar product, it satisfies the following parallelogram identity
|v + w|2 + |v - w|2 = 2|v|2 + 2|w|2
For the l4 norm, consider two different vectors which have coordinates all equal to zero but one which is equal to 1. Then we have
|v + w|2 + |v - w|2 = 2(2)
and
2|v|2 + 2|w|2 = 4.
Since these numbers are different, the parallelogram identity fails and so l4 is not a Hilbert space.
2. (a) If the space X has dimension n e N, we can choose a basis ej, j = 1, .., n and for any vector v in the space we have a representation
v = vjej .
j=1,..,n
Then, we have
lvlp < max lejlp IvjI
j=1,..,n
On the other hand, the function
(vj)j=1,..,n → l vjejl
j=1,..,n
attains its minimum on the closed bounded set defined by the normalisation condition j=1,..,nIvjI = 1. This yields
lvlp 2 mp IvjI.
j=1,..,n
with some mp e (0, o).
Since similar property holds with the p˜ norms, we conclude that the norms are equivalent.
Next we give examples of infinite dimensional proper closed subspaces on which both norms are well defined, but not equivalent.
Let
Vj = {(zk)keN e lp : zj = 0}
for any j e N. Vj are proper closed subspaces of lp and infinite dimensional (with a canonical basis (δk . ), k j, respectively). It contains truncated sequences from the space lp˜ of the form
(zkχL (k))keN
where χL (k) = 1 for k < L and zero otherwise. One can choose a sequence which is Cauchy with respect to lp˜ norm for p˜ > p, but not convergent with respect to lp norm. Thus the norms are not equivalent.
(b) Using Hahn-Banach theorem one can show that the dual space of c([0, 1]) is given by Rieman-Stielties integrals
c([0, 1]) 3 f → fdF
associated to functions F of finite variation on [0, 1]. Consider a set of functions χx = χ [x,1], which for x e (0, 1) are equal to zero on [0, x) and one otherwise. This is an uncountable set of linearly independent functions with the property that total variation norm of a difference of any of two such functions is equal to one. This implies that the dual space cannot be separable.
3, B |
(c) Since the subspace W is closed, it is a Hilbert space in itself. Then, by Riesz
representation theorem there exists a unique vector u之 e W such that
z(v) =(u之 , v)
and
lzl = lu之 l.
By Hahn-Banach theorem, there exists a functional
on l2 such that its restriction to W coincides with z and its norm satisfies l
l = lzl. Again invoking Riesz representation theorem on l2 , we can find a unique vector
之 e l2 representing
and such that l
之 l = l
l = lzl = lu之 l. Since by our assumption
(u之 , w)=(之 , w)
for all w e W , the vector 之 - u之 is orthogonal to W and in particular to u之 . By parallelogram identity we have
l之 + u之 l2 + l
之 - u之 l2 = 2lu之 l2 + 2l
之 l2
and hence, using 之 - u之 l u之 , we get
l之 + u之 l2 + l
之 - u之 l2 = l
之 - u之 + 2u之 l2 + l
之 - u之 l2
= 2l之 - u之 l2 + 4lu之 l2 .
This together with parallelogram identity implies
l之 - u之 l2 = l
之 l - lu之 l = 0.
Thus we conclude that
之 = u之 .
…← (a) We have
lTnf(x) _ Tngl2 = and hence
│ │ 2
|ξ|2 Hn(x)2 │ Hn(y) (f(y) _ g(y)) µ(dy)│ µ(dx).
│ │
lTnf _ Tngl = |ξ| . lf _ gl.
Thus the operator is strictly contractive for |ξ| < 1.
(b) To show that for |ξ| = 1 it is only weakly contractive (or to show that it is not contractive at all for |ξ| > 1 ), consider f = Hn and g = λHn with λ e _. Using orthogonality of Hermite polynomials we can show that
Tn(αHk + βHn) = Hk + βξHn .
Hence we get the following condition for a fixed point
α = 1, β = βξ.
If |ξ| < 1, the only solution is α = 1, β = 0, while for |ξ| = 1 one gets infinitely solutions with α = 1, β e _.
(c) For n e N define
n
unseen ψ |
We have
n 1 │ │ 2 n 1
n=2 pn(2) │ │ n=2 pn(2)
Thus KN is bounded. This is a finite rank operator mapping l2 into a finite dimensional subspace spanned by Hn, n < N . Since a bounded operator with finite dimensional range is compact, each Kn is a compact operator. We can show that the sequence KN, N e N of compact operators converges in Hilbert-Schmidt norm to K as follows. First of all we have
lKN _ KlH(2) >S = |IHn , (KN _ K)Hn|2
n-Nì2
1
=
n-Nì2 pn(2) .
It is known that the k-th prime number asymptotically is of order
pk ~ k log(k),
and from basic series analysis one knows that
< o.
This implies that
lKN _ KlH(2) >S → 0.
Noo
Since the Hilbert-Schmidt norm dominates the operator norm, so KN → K in the operator norm as well. This together with the fact that KN’s are compact implies that K is a compact operator.
4. (a) By definition of the subspace
V = {f é √ ([0, 1]) : 刁c é 佟 Ⅴx é [1/3, 2/3] f(x) = c}
if fn é V, n é …, converges to a function f in the supremum norm, then it converges pointwise uniformly on a compact interval [0, 1]. Hence the limit f(x) = limnì&fn(x) is well defined and it is a continuous function. Moreover for any x é [1/3, 2/3] we have fn(x) = cn and the sequence of numbers cn , n é …, is
convergent to a number c é 佟, the same for all x é [1/3, 2/3]. Hence f é V and so the subspace V is closed.
(b) We need to show that V does not contain any open ball of (√ ([0, 1]), |.|g), i.e. that any set
Bf,e = {g é √ ([0, 1]) : |g - f| < ε}
contains functions which do not belong to V . To this end consider a function
,.f(x), ī≥ x é [0, 1/3] u [2/3, 1]
.
.
..3ε(1/2 - x) + f(2/3) + ε/2 ī≥ x é [1/3, 1/2]
By our definition of the function g, we have
|g -f| = sup Ig(x) -f(x)I = sup Ig(x) -f(x)I = Ig(1/2) -f(1/2)I = ε/2. uà|à ,|J uà||/á ,今/áJ
Thus g é Bf,e and g
V . Since ε é (0, &) was arbitrary, we conclude that the closed set V does not contain any open ball and so its interior is empty.
(c) The proof is based on use of Banach-Steinhaus theorem: Let X and Y be a Banach space and a normed space, respectively. If a family of operators Ta : X → Y , α é ← , for some index set ←, satisfies
Ⅴx é X刁cu é (0, &)
then
sup |Tax|Y ≤ cu ,
aà←
sup |Ta| < &.
aà←
We assume that for some (zj é 佟)jà… we have
Ⅴx é lp(ρ)
zjxjρj < &.
j
Consider a family of functionals (fn)nà…
n
fn(x) = zjxjρj .
j=|
Using H¨older inequality, with +
= 1, we have
z
Ifn(x)I ≤ l n IzjIaρj、 J |x|l尹 ←o(,
and thus all operators 1n are bounded from the Banach space ap(o) to ← (which with the modulus l . l form a normed space). Moreover
1
|1n| < l3jlgoj. g .
In fact one can choose z so that the equality is attained. By our assumption of convergence of the series
3jzjoj
j
we have
Vz e ap 3cz e (0, o)
Hence by Banach-Steinhaus theorem
sup l1n(z)l < cz
n
sup sup l1n(z)l < o.
n lzllp(o)
That is using our expression for the norm |1n| we have
1
╱ n l3jlgoj、 g < o,
n
which implies that 3 e ag(o). Finally since o is normalised, we can use H¨older inequality
|3 |lg~(o) < |3 |lg(o)
for all g˜ e [1, g].
5. (a) Suppose (X, | _ |X ) and (Y, | _ |Y ) are Banach spaces such that X < Y and aC e (0, o) Vx e X |x|Y = C|x|X
the identity map from X to Y is a compact operator (i.e. maps bounded sets in X to sets in Y which after closure are compact). Then we say that the Banach space (X, | _ |X ) is compactly embeded into the Banach space (Y, | _ |Y ).
(b) For a nonnegative integer m e Z+ and p e [1, o], and for an open set Ω < RN , the space Wm,p(Ω) is by definition the space of (equivalence classes of) functions f e Lp(Ω) such that Vα f e Lp(Ω) for all derivations Vα of length |α| = m, that is for any ϕ e c< compactly supported in Ω we have
fVα ϕdλ = (l1)lαl gϕdλ
and g e Lp(Ω). It is a normed space equipped with the norm
|Vα f |Lp(Ω)
lαl_m
or equivalent norm, for p e [1, o)
1
|f |m,p = ╱
while for p = o
lαl_m
unseen c |
(c) First of all we note that
(y _x)2
VTf (x) . (x l y)e_ 2 f (y)λ(dy)
Ω
and hence for any r, s e (0, o)
|VTf |p = ╱ Ω │ Ω (x l y)e_ f (y)λ(dy)│p λ(dx)、
(w)2
= |Ω|||w|e_ 2 |s _ |f |r
Hence for f e Lr, we have Tf e W1,p, for any p = r Using this, in particular with 2 = r, together with the Relich-Kodrachov compact embedding theorem we conclude that
{Tf : |f |2 = 2}
is compact in L2(Ω).