ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

4 ANALYSIS QUALS
X fndν → a. Let gn = max{f1, . . . , fn}. Then gn ∈ F for all n. Furthermore,
gn increases pointwise to f = sup{fn}. Then
X gndν → a, and by
the monotone convergence theorem, a =
fdν and f ∈ F.
X
Define dµs = dµ − fdν, and dµa
= fdν. Then if ν(E) = 0, we have that
E fdν = 0, so µa 0 and E ∈ M such that ν(E) = 0
and E is a positive set for µs−εν. That is, µs ≥ εν on E. But then εχEdν ≤
χEdµs, so εχEdν ≤ χE(dµ−fdν), which implies that (εχE +f)dν ≤ µ(E).
But this contradicts our selection of a as the supremum of all such integrals.
Therefore, we conclude that µs ⊥ ν. This completes the proof.
Problem 7: Prove Gorsat’s theorem: If f : C → C is complex differentiable
(and so continuous), then for every triangle ∆ ⊂ C
f(z)dz = 0
∂∆
where line integral is over the three sides of the triangle.
Solution. Let ∆ be a closed triangle in C and let a, b, c be the vertices of ∆.
Let a ′ , b ′ , c ′ be the midpoints of [b, c], [a, c], [a, b], respectively. Consider the
four triangles ∆ j , j = 1, 2, 3, 4, with vertices {a, c ′ , b ′ }, {b, a ′ , c ′ }, {c, b ′ , a ′ },
and {a ′ , b ′ , c ′ }. Let
Then
J =
J =
Hence,
∂∆
4
j=1
∂∆ j
f(z)dz.
f(z)dz.
f(z)dz
≥ |J/4|
∂∆k for some k. Define ∆1 to be this ∆k . Replace ∆ with ∆1 and repeat. This
yields a sequence of triangles {∆n} such that
∆ ⊃ ∆1 ⊃ ∆2 ⊃ . . . ⊃ ∆n ⊃ ∆n+1 ⊃ . . .
and the length of ∂∆n is 2−nL, where L is the length of ∂∆. Thus, we have
that
|J| ≤ 4 n
f(z)dz
for all n.
∂∆n
Choose z0 ∈ ∞
n=1 ∆n, and we note that in fact this z0 is unique. Let
ε > 0. As f is complex differentiable, we may choose r > 0 such that if
|z − z0| < r, then
for z ∈ ∆.
|f(z) − f(z0) − f ′ (z0)(z − z0)| ≤ ε |z − z0|
Choose n such that |z − z0| < r for all z ∈ ∆n. Note also that |z − z0| ≤
2 −n L for all z ∈ ∆n.
We use the fact that
γ zn dz = 0 for any closed path γ ⊂ C and n =