ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

so y ∈ B(x, ε), so y ∈ B(x (n1)
i , 1
n1
B(x (2N)
k
, 1
2N
) ⊂ B(x(n1)
i
ANALYSIS QUALS 35
, 1
n1
) ∪ B(x(n2)
j
) ∪ B(x(n2) j , 1 ). Thus, we have that
n2
, 1
n2
). Hence, B is a base for the
topology, so X is second countable.
(c) Let {x1, x2, . . .} be the dense subset of X found in (a). By Urysohn’s
Lemma, for n, m ≥ 1, there exists ψn,m ∈ C(X) such that ψn,m = 1
on B(xn, 1
m ) and ψn,m is supported on B(xn, 2
Then {ψn,m} separates points, as the balls are a base for the topology
by (b). By the Stone-Weierstrass theorem, the algebra of rational
polynomial combinations of ψn,m is dense, so C(X) is separable.
Problem 4: Let f, g ∈ L2 (R) be two square-integrable functions on R (with
the usual Lebesgue measure). Show that the convolution
f ∗ g(x) := f(y)g(x − y)dy
of f and g is a bounded continuous function on R.
Solution. First, we show it is bounded:
|f ∗ g(x)| =
f(y)g(x − y)dy
R
1/2
≤
R
|f(y)|
R
2 dy
= f2 g2 .
m ).
|g(x − y)|
R
2 1/2 dy
Hence, f ∗ g is bounded. Now we show that f ∗ g is continuous.
|f ∗ g(x1) − f ∗ g(x2)| = f(y)g(x1
− y)dy − f(y)g(x2 − y)dy
R
R
=
f(y) [g(x1 − y) − g(x2 − y)] dy
R
= f(x1 + y) [g(y) − g(x2 − x1 − y)] dy
R
≤ f
2 g − g(x2−x1) ,
2
where g (x2−x1) is the translation. But
g − g(x2−x1) → 0 as |x2 − x1| →
2
0. Hence f ∗ g is continuous.
Problem 5: Let H be a Hilbert space, and let T : H → H be a bounded
linear operator on H.
(a) Show that if the operator norm T of T is strictly less than 1, then
the operator 1 − T is invertible.
(b) Let σ(T ) denote the set of all complex numbers z such that T − zI is
not invertible. (This set is known as the spectrum of T .) Show that
σ(T ) is a compact subset of C.