ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

46 ANALYSIS QUALS
γR
Solution. The problem is false as stated, as some assumption of boundedness
must be made. A counterexample is to let X be l ∞ , the space of
bounded sequences (x1, . . . , xn, . . .), and let
A : (x1, . . . , xn, . . .) ↦→ (x1, x2/2, . . . , xn/n, . . .).
Then A is linear and surjective, so 0 ∈ ρ(A), but 1/n ∈ ρ(A) for any n ∈ N,
so ρ(A) is not open.
That said, we alter the problem and change the definition of ρ(A). Let
ρ(A) = {λ ∈ C : (λ − A) −1 exists and is bounded}
Let G be the set of operators on X with bounded inverse. We claim that
G is open. Let S ∈ G and suppose that T − S < S −1 1 .
Then S −1 T − I = S −1 (T − S) < 1, so we have that
(S −1 T ) −1 = (I − (I − S −1 T )) −1 = (I − S −1 T ) i exists.
Therefore T −1 = (S −1 T ) −1 S −1 , so T is invertible. Therefore, G is open.
Define f : C → B(X) by α ↦→ (αI − T ). Then f is continuous, so f −1 (G)
is open. But f −1 (G) = ρ(A), so ρ(A) is open.
Problem 8: Determine the number of zeros of the polynomial
p(z) = z 4 + z 3 + 4z 2 + 2z + 3
in the right half-plane {z : Re z > 0}.
Solution. Let ΓR be the contour that is the right half of the circle of radius
R centered at the origin. Let γR be the semi-circle part of the contour, and
let γR ˆ be the segment from Ri to −Ri.
Note that on {Re z = 0}, we have that Re p(it) = t 4 − 4t 2 + 3 = (t 2 −
3)(t 2 − 1), and Im p(it) = −t 3 + 2t = −t(t 2 − 2). Thus, there are no zeros
on {Re z = 0}.
Thus, we can use the argument principle in this contour, which says that
p
ΓR
′ (z)
p(z) dz = 2πi(number of zeros in the interior of ΓR). By taking R → ∞,
we can calculate the total number of zeros in {z : Re z > 0}.
On γR, p(z) = z 4 + O(|z| 3 ), and p ′ (z) = 4z 3 + O(|z| 2 ), so p′ (z)
p(z) = 4z−1 +
O(|z| −2 ). Thus,
p ′ (z)
dz =
p(z)
π
0
Ri · 4 · e−θ
R dθ + O(|R|−1 ) = 4πi + O(|R| −1 ) → 4πi as R → ∞.
On ˆ
γR, note that critical points (points where p crosses either the real or
imaginary axes) are at t = 0, ±1, ± √ 2, ± √ 3 for z = it. Suppose R > √ 3.
Then on ( √ 3i, Ri), p takes values in the 4th quadrant (by noticing that
the real part is positive and the imaginary part is negative). Similarly, we