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ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

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<strong>ANALYSIS</strong> QUALS 25<br />

b) I present the proof in the (simple) case in which H is a separable<br />

Hilbert space (which is how Stein-Shakarchi defines their Hilbert spaces).<br />

In this case, let {en} be an orthonormal basis for H.<br />

Define Tn(y) = 1<br />

y 〈en, y〉 y. Then<br />

Tn(y) ≤ y , and Tn(en) = 1, so Tn = 1.<br />

But by Bessel’s inequality,<br />

<br />

n<br />

|〈en, y〉| 2 ≤ y 2 , so 〈en, y〉 → 0 for all y.<br />

Thus, lim Tn(y) = 0 for all y ∈ H.<br />

Problem 3: Let X be a Banach space and let X ∗ be its dual Banach space.<br />

Prove that if X ∗ is separable the X is separable.<br />

Solution. Let {fn} be a countable dense set of X ∗ . For each n, choose<br />

xn ∈ X such that xn = 1 and |fn(xn)| ≥ 1<br />

2 fn.<br />

Let S be the set of all linear combinations of the xn with rational coefficients<br />

(rational real and imaginary parts if X is a Banach space over C).<br />

Thus, S is countable. We aim to show that S is dense in X.<br />

Let Y = S and assume for sake of contradiction that Y = X. Then by<br />

Hahn-Banach theorem, we may choose f ∈ X∗ such that f(y) = 0 for all<br />

y ∈ Y and f = 0. Then by density of {fn}, choose a sequence fin → f.<br />

Then<br />

1<br />

f − fin ≥ |(f − fin )(xin )| = |fin (xin )| ≥ fin 2<br />

for all n ∈ N. But then as fin → f, we must have fin → 0, which means<br />

that f = 0, a contradiction. Thus Y = X. <br />

Problem 4: Let f(x) be a non-decreasing function on [0, 1]. You may assume<br />

the theorem that f is differentiable almost everywhere.<br />

a) Prove that 1<br />

0 f ′ (x)dx ≤ f(1) − f(0).<br />

b) Let {fn} be a sequence of non-decreasing functions on the unit interval<br />

[0, 1], such that the series F (x) = ∞<br />

n=1 fn(x) converges for all x ∈<br />

[a, b]. Prove that F ′ (x) = ∞<br />

n=1 f ′ n(x) almost everywhere on [0, 1].<br />

Proof. a) Define the difference quotient<br />

gn(x) =<br />

f(x + 1/n) − f(x)<br />

,<br />

1/n<br />

and then by the theorem we are permitted to assume, gn(x) → f ′ (x)<br />

as n → ∞ for a.e. x. As f is non-decreasing, we have that gn is<br />

non-negative, so by Fatou, we have that<br />

1<br />

0<br />

f ′ (x)dx ≤ lim inf<br />

1<br />

0<br />

gn(x)dx.

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