## Qual Attempt #1

*Date Written: July 14, 2023; Last Modified: July 14, 2023*

Practise makes perfect, and we are far from perfect.

I’ll be making posts like this every time I attempt a (past) qualifying exam on my own time. In these, I’ll post a copy of the exam (all of which can be found on UCLA’s website, by the way), my solutions that I completed within a 4-hour time frame, alongside corrections to mistakes, additional solutions that I thought of after time was up, and other commentary.

This time, I only had a 3-hour opening to work with, hence my performance was rather lacking. I only completed 4 problems (two from each section), and I made an error in computing a contour integral. Here are the exam problems and my solutions.

Overall, I can make two broad statements about my performance:

- My real analysis is weak, probably because I’ve only been self-studying. This means that doing these qualifying exams is for the better.
- My complex analysis is rusty. I haven’t reviewed my class notes from over a year ago, and it definitely would have helped for at least two problems. I’ll hopefully try another exam next week, but it’s hard to say with travel plans and family obligations looming on the horizon.

Here are some problems that I have something to say about. Not every problem I solved is here, and I did not solve every problem here.

**Problem 1. Egorov's Theorem**

Prove Egorov's theorem, that is: conisder a sequence of measurable functions \(f_n:[0, 1]\to\mathbb R\) that converge (Lebesgue) almost everywhere to a measurable function \(f:[0, 1]\to\mathbb R\). Then for any \(\epsilon>0\) there exists a measurable set \(E\subset [0, 1]\) with measure \(\lvert E\rvert<\epsilon\) such that \(f_n\) converge uniformly on \([0, 1]\setminus E\). Hint: You may want to consider the sets \[E_n(k)=\bigcap_{m\geq n}\left\lbrace x:\left| f_n(x)-f(x)\right| <\frac 1k\right\rbrace.\]

This was perhaps one of the easier problems, and I was able to solve it without a problem. The proof, which I believe is standard, can be found in Folland’s textbook and probably any standard real analysis textbook.

**Problem 4. **

Let \(f\in C^\infty\left([0,\infty)\times[0, 1]\right)\) such that \[\int_0^\infty \int_0^1\left| \partial_t f(t, x)\right|^2\left(1+t^2\right) dx dt < \infty. \] Prove that there exists a function \(g\in L^2([0, 1])\) such that \(f(t, \cdot)\) converges to \(g(\cdot)\) in \(L^2([0, 1])\) as \(t\to\infty\).

I looked at this problem and immediately skipped it because I got scared by the partial derivative. Perhaps the first choice to take is \[g(x)=\int_0^\infty \partial_t f(t, x) dt+f(0, x),\] to which \(f(t,\cdot)\) converges to by the fundamental theorem of calculus. However, it’s hard for me to justify the convergence of this integral, and perhaps there’s a better choice of \(g\) for this reason.

**Problem 5. **

For a function \(f:\mathbb R\to\mathbb R\) belonging to \(L^1(\mathbb R)\), we define the Hardy-Littlewood maximal function as follows: \[(Mf)(x) := \sup_{h>0} \frac{1}{2h} \int_{x-h}^{x+h}\left| f(y)\right| dy.\] Prove that it has the following property: There exists a constant \(A\) such that for any \(\lambda>0\), \[\left|\left\lbrace x\in\mathbb R : (Mf)(x) > \lambda\right\rbrace\right| \leq \frac A\lambda \lVert f\rVert_1,\] where \(\left| E\right|\) denotes the Lebesgue measure of \(E\). If you use a covering lemma, you should prove it.

Intuitively, multiplying the left side by \(\lambda\) should give you something less than \(\lVert f\rVert_1\); the set on the left is roughly the region where \(f(x)\) locally has an “average value” of at least \(\lambda\). However, the obstruction I faced when attempting this arose from the potentially overlapping limits of integration; it seems that this is what the “covering lemma” is needed for, and it’s the information captured by the coefficient \(A\). Sadly, my brain is not covered by lemmas.

**Problem 7. **

Compute \[\int_0^\infty \frac{\cos x}{\left(1+x^2\right)^2}dx.\] Justify all steps.

Not a difficult problem, but I forgot to put the derivative in the computation of the residue. I was thoroughly confused by why my answer was negative. I moved on instead of crying about it.

Note here that whenever a problem involving \(\cos x\) or \(\sin x\) in an integral arises, it’s almost always better to express them as real or imaginary components of \(\exp(ix)\). This lets you do contouring, since \(\exp(iz)\) decays in the upper half-plane while both \(\cos z\) and \(\sin z\) grow exponentially in both imaginary directions. This was a trick I learned back in my first quarter of complex analysis, so I’m glad I haven’t lost everything.

**Problem 8. **

Determine the number of solutions to \[z-2-e^{-z}=0\] with \(z\) in the right half-plane \(H=\left\lbrace z\in\mathbb C : \Re(z)>0\right\rbrace\).

I attempted to use the argument principle here. I stupidly wound up spending the better part of an hour trying to refine my estimates in the argument principle, both by getting better point estimates of my integrand and by picking a better contour, but to no avail. I determined that there were either 1 or 3 solutions, but I couldn’t figure out which of the two there were.

Rather, the right thing to do was to use RouchÃ©’s theorem. With \(f(z)=z-2-e^{-z}\), we have \(f(z)=0\) implies \(z=2+e^{-z}\) implies \(|z|<3\) when \(z\in H\): \(\left| e^{-z}\right|=e^{-\Re z}<1\) since \(\Re z>0\). Hence, all of the zeroes of \(f\) occur in the region \(K=H\cap\left\lbrace z\in\mathbb C: |z|<4\right\rbrace\). We need the extra wiggle room of \(|z|<4\) for the soon-to-happen application of the theorem.

Consider \(g(z)=e^{-z}\). We claim \(|g(z)|\leq |f(z)|\) on \(\partial K\). The boundary of \(K\) comes in two pieces, which we’ll handle separately. First, there’s \(\left\lbrace z=yi : -4\leq y\leq 4\right\rbrace\). Here, we note that \[|f(z)|\geq |z-2| - \left|e^{-z}\right| > 2-1=1.\] We have \(|z-2|>2\) because \(z\) is purely imaginary in this segment. On the other hand, \(|g(z)|\leq 1\) everywhere, so \(|g(z)|\leq |f(z)|\) here.

The other segment of \(\partial K\) is \(\left\lbrace z\in H : |z|=4\right\rbrace\). We instead get that \[|f(z)|\geq |z|-2-\left|e^{-z}\right| = 1,\] so we conclude \(|g(z)|\leq |f(z)|\) everywhere on \(\partial K\). Thus, \(f(z)\) and \(f(z)+g(z)\) have the same number of zeroes in \(K\). However \(f(z)+g(z)=z-2\), which has exactly one zero in \(K\). Hence, \(f\) has exactly on zero in \(K\); as all of its zeroes in \(H\) are in \(K\), it has a total of one zero in \(H\).

**Problem 10. **

Let \(\Omega\subsetneq \mathbb C\) be a simply connected domain, and \(f:\Omega\to\Omega\) be a holomorphic mapping. Suppose that there exist points \(z_1,z_2\in\Omega\), \(z_1\neq z_2\), with \(f\left(z_1\right)=z_1\) and \(f\left(z_2\right\)=z_2\). Show that \(f\) is the identity mapping on \(\Omega\).

This is where my rustiness with complex analysis hurt me the most. I was able to reduce to the case where \(\Omega=\mathbb D\) and one of the fixed points was the origin. I distinctly remember studying functions on the unit disc, but I unfortunately forgot just about everything that could have been helpful.

**Problem 12. **

A holomorphic function \(f:\mathbb C\to\mathbb C\) is said to be of exponential type if there are constants \(c_1, c_2>0\) such that \(|f(z)|\leq c_1 e^{c_2 |z|}\) for all \(z\in\mathbb C\). Show that \(f\) is of exponential type if and only if \(f'\) is of exponential type.

I just didn’t have time to do this, but this seems on the easier side (fortunately). I believe the “if” direction follows immediately by integrating on a radial line from \(0\) to \(z\), then using the triangle inequality on the line integral. The “only if” direction should follow from the higher-order Cauchy integral formula. In particular, we have \[f’(z)=\frac{1}{2\pi i} \oint \frac{f(w)}{(w-z)^2} dw,\] where the contour integral is taken over a circle of radius \(|z|+1\) centred at the origin. Using the crudest bound on the integrand’s magnitude is enough: we have \(|w-z|^2 \geq 1\), so we get the bound \[ \lvert f’(z)\rvert \leq \frac{|z|+1}{2\pi i} \cdot c_1\exp\left(c_2(|z|+1)\right).\] It is then not hard to show that \(f’\) is of exponential type from here.