Download Functional analysis exercises 02

(October 9, 2012)
Functional analysis exercises 02
Paul Garrett [email protected] http://www.math.umn.edu/egarrett/
[This document is http://www.math.umn.edu/˜garrett/m/fun/exercises 2012-13/fun-ex-10-05-2012.pdf]
Due Wed, 24 Oct 2012, preferably as PDF emailed to me.
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[02.1] Convincingly and not-ugly-ly prove that e−1/x (naturally extended by 0 at x = 0) is infinitely
differentiable at 0.
[02.2] Show that for 0 ≤ cn ∈ R with cn decreasing monotonicly to 0, the Fourier series
P
n cn
einx converges
at x 6∈ 2πZ, although not necessarily absolutely.
[02.3] (Fejér kernel for S 1 × S 1 ) Prove completeness of suitable exponentials in L2 (S 1 × S 1 ) or higher
dimensions directly, by constructing an approximate identity consisting of finite Fourier series. As expected,
consider the square of the corresponding Dirichlet kernels
X
X
X X
eimx+iny =
eimx ·
einy
(Dirichlet kernel)
DM,N (x, y) =
|m|≤M |n|≤N
|m|≤M
|n|≤N
The Fejér kernel is the normalization
ΦM,N (x, y) = Z
DM,N (x, y)2
(Fejér kernel)
DM,N (u, v)2 du dv
S 1 ×S 1
[02.4] Map Cn → Cn+1 by (z1 , . . . , zn ) → (z1 , . . . , zn , 0). Let V = colimn CnS=
Cn , with the colimit
topology, in which a basis of opens at 0 is given by convex hulls of unions B = n Bn where Bn is an open
ball of some positive radius, at 0, in Cn . Show that V violates the conclusion of the Baire category theorem,
so is not complete-metrizable. Here Cauchy sequences {xn } are those such that, given a neighborhood N of
0, there is no such that xm − xn ∈ N for all m, n ≥ no . Show that Cauchy sequences converge.
S
n
2
d
2
2
1
[02.5] Show that for complex w the equation ( dx
2 − w ) uw = 0 has a solution in C (S ) only when w ∈ iZ.
Let δ per be the 2πZ periodic Dirac δ-function, the continuous linear functional on C o (R/2πZ) given by
δ per f = f (0). With w ∈ C and w 6∈ iZ, solve
d2
− w2 uw = δ per
2
dx
for uw on R/2πZ. (Hint: expand δ per in a Fourier series.) Identify the residues of the L2 (S 1 )-valued
meromorphic function w → uw .
[02.6] Show that there are no eigenvectors for the Volterra operator T : L2 [0, 1] → L2 [0, 1] given by
Z
T f (x) =
x
f (y) dy
0
d
By design, dx
T f = f for f ∈ C o [0, 1]. Show that (T − λ)u = f is solvable for u when 0 6= λ ∈ C and
1
f ∈ C [0, 1] with f (0) = 0, and the solution is unique. With T ∗ the Hilbert-space adjoint of T , show that
Z 1
Z 1
∗
∗
(T T f )(x) =
min (x, y) · f (y) dy
(T T f )(x) =
1 − max(x, y) · f (y) dy
0
∗
0
∗
Find eigenvectors for T T and T T .
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