ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 29
Problem 2: Is every vector space isomorphic as a vector space to some Banach
space? Prove your answer. (Banach space = complete normed vector
space, as usual).
Solution. No. Let V be the vector space consisting of all infinite sequences
(over R, for example) with only finitely many non-zero terms. Assume for
sake of contradiction that V can be given a Banach space structure.
Let Vn be the subspace of V consisting of the sequences whose components
are all zero after the nth. Then
∞
V = Vn.
n=1
Note that Vn is finite dimensional, so it is isomorphic as a Banach space
to R n , and thus is a closed subspace. However, Vn is nowhere dense, as
Vn contains no open balls in V . Thus, by the Baire Category Theorem,
Vn = V , a contradiction.
Problem 3: Prove: If f : [0, 1] → R is an arbitrary function, not necessarily
measurable, then the set of points at which f is continuous is a Lebesguemeasurable
set.
Proof. Let
Sx(δ) = sup{|f(x1) − f(x2)| : x1, x2 ∈ (x − δ, x + delta)},
φ(x) = inf{Sx(δ) : δ > 0}.
Note that
1
1
Sx ≥ Sx
k k + 1
for every k ∈ N. We claim that f is continuous at x if and only if φ(x) = 0.
First suppose that φ(x) = 0. Let ε > 0 and choose δ > 0 such that
Sx(δ) < ε. Thus, for any y ∈ (x − δ, x + δ), we have that |f(y) − f(x)| < ε,
so f is continuous at x.
Now assuming f is continuous at x, we have that for all ε > 0 we may choose
δ such that |x − y| < δ implies |f(x) − f(y)| < ε, and hence Sx(δ) < 2ε.
Thus, φ(x) = 0.
Next, we claim that {x : φ(x) < α} is an open set for every α (that is, φ is
upper-semicontinuous). Suppose φ(x) < α and choose N such that
1
Sx < α whenever k > N.
k
Suppose k > N. Let y ∈ x − 1
1
1 1
2k
, x
+ 2k
. Note
that then
y − 2k , y + 2k
⊂
1
1
1 1
(x−1/k, x+1/k), so φ(y) ≤ Sy 2k ≤ Sx k < α. Thus, x − 2k , x + 2k ⊂
{x : φ(x) < α}, so the set is open.
But then we have that
∞
{x : φ(x) = 0} = {x : φ(x) < 1
k },
k=1
and thus the set of points of continuity of f is a Gδ set, and hence is
measurable.