ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

44 ANALYSIS QUALS
(3) Show that the only f ∈ L 1 (R) for which f ∗ f = f is f = 0.
Solution. (1) We have that
Ag(f) =
f(x − y)g(y)dydx
≤ f1 g1 ,
R
R
so as g ∈ L1 , we have that Ag is a bounded operator.
(2) By part (1), we have that Ag ≤ g. We have that
Ag(g) = g(x−y)g(y)dydx = g(y) g(x−y)dxdy = g(y) g dy = g 2 ,
R
R
R
R
and so Ag ≥ g, and hence Ag = g.
(3) Suppose that f ∗ f = f. Then by taking Fourier transforms, we have
that f ˆ∗
f = ˆ f. But using properties of Fourier transforms, we have
f ˆ∗
f = ˆ f · ˆ f. Thus, ˆ f 2 − ˆ f = 0 a.e. Hence ˆ f = 0 or 1 a.e.
But since f ∈ L1 , we have that ˆ f is continuous and vanishes at ∞ by
Riemann-Lebesgue theorem, so we conclude that ˆ f = 0 everywhere.
Then by Fourier inversion, we conclude that f = 0.
Problem 4: Let α ∈ R\Q and let T : L 2 ([0, 1]) → L 2 ([0, 1]) be defined by
(T f)(x) = f(x + α mod 1).
Denote Snf = f + T f + T 2f + · · · + T n−1f. Do the following:
(1) For any f ∈ L2 ([0, 1]), prove that 1
nSnf converges in L2 . Identify the
limit.
(2) Suppose f : [0, 1] → R is continuous with f(1) = f(0). Show that the
convergence in (1) is uniform.
Solution. We prove both at once. We claim that 1
nSnf(x) → 1
f(t)dt for
0
every x ∈ [0, 1], and that the convergence is uniform for continuous periodic
functions.
First, let f(x) = e2πikx . If k = 0, the statement is clearly true. Suppose
k = 0. Then the statement is that
N−1
1
e
N
2πik(x+nα) →
n=0
1
0
e 2πikt dt.
But by the geometric series, the LHS is equal to
2πikx 1 1 − e
e
N
2πiNαk
1 − e2πiαk and since α /∈ Q, the denominator is never 0. Hence, this approaches 0 as
N → ∞, and 1
0 e2πikxdx = 0, so the statement is true for all trigonometric
polynomials.
Now let ε > 0, f a continuous function from R/Z to R. As trigonometric
polynomials are uniformly dense in the space of continuous periodic
functions, we may choose P (x) trigonometric such that
sup |f(x) − P (x)| <
x∈R/Z
ε
3 .
R