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Revisiting Qual Attempt #1

Date Written: July 28, 2023; Last Modified: July 28, 2023

I previously attempted to work on a past qualifying exam in analysis with rather dismal results. In the previous post, I only gave solutions for 6 of the 12 problems (2 real analysis, 4 complex analysis). Over the past few days, I have been able to put together solutions for 5 of the remaining problems, with only one problem from the real analysis section left unsolved. I do have some thoughts about it, but I’m happy with solving 11 of the 12 problems (albeit taking far longer than should be necessary).

This is the problem that I had to leave completely unsolved. It looks like it’s related to Fourier analysis or harmonic analysis, which I have yet to read up on (I plan to do so later this summer). I haven’t attempted to compute the first integral; things like the divergence theorem come to mind, but it seems difficult to apply to this situation. For the second part, this looks like a routine application of Hahn-Banach, but upon closer inspection, one may instead use the fact that $C^\infty$ is dense in $L^2\left(\R^3\right)$ together with continuity of this map (with respect to the $L^2$ norm). However, the details are hazy at best, and I’m not clear on how the first integral is relevant. Oh well.

For the first part, we may consider that for any $h\in\mathbb R^3$, $$\begin{align*} \left\lvert f*g(x+h)-f*g(x)\right\rvert &\leq \int_{\mathbb R^3}\left\lvert\left(f(x+h-y)-f(x-y)\right) g(y)\right\rvert dy \\ &\leq \left\lVert f(x+h-\cdot)-f(x-\cdot)\right\rVert_p \left\lVert g\right\rVert_q. \end{align*}$$ By the continuity of translations on $L^p\left(\mathbb R^3\right)$, it follows that the first multiplicand vanishes as $h\to 0$, as desired.

Frankly, the second part still eludes me. I believe there’s an approximation argument to be made — it seems that the claim should be pretty clear when either $f$ or $g$ is compactly supported. However, the specifics of this type of argument seem difficult to justify, as there will be two limits to deal with: one limit for approximating $f$ or $g$, and another for $\lvert x\rvert\to\infty$.

It turns out that my intuition about this problem was correct. By Tonelli’s theorem, we have that $$\int_0^1\int_0^\infty \lvert\partial_t f(t, x)\rvert^2 \left(1+t^2\right) dt dx < \infty.$$ In particular, the inner integral converges for almost every $x$. Then, by Hölder’s inequality, we have: $$\int_0^\infty \lvert\partial_t f(t, x)\rvert dt \leq \left\lVert \partial_t f(t, x) \sqrt{1+t^2}\right\rVert_{L^2([0,\infty))}\cdot \left\lVert \left(1+t^2\right)^{-\frac 12}\right\rVert_{L^2([0,\infty))}.$$ Both of the norms on the right are finite: the first multiplicand is finite by the first remark, and the second is just the squareroot of $\int_0^\infty \frac{1}{1+t^2} dt =\frac\pi 2$. Hence, the following function is well-defined for almost every $x$: $$g(x)=f(0, x)+\int_0^\infty \partial_t f(t, x) dt,$$ and this is equal to $\lim_{t\to\infty} f(t)$ by the fundamental theorem of calculus. Moreover, we have by Minkowski’s inequality that $$\lVert g(\cdot) \rVert_2\leq \lVert f(0, \cdot)\rVert_2 + \left\lVert\int_0^\infty \partial_t f(t, \cdot) dt\right\rVert_2.$$ $f(0, x)$ is continuous on a compact set, so it’s in $L^2$. For the second term in the bound, we have $$\left\lVert \int_0^\infty \partial_t f(t,\cdot) dt\right\rVert_2^2 = \int_0^1 \int_0^\infty \lvert\partial_t f(t, x)\rvert^2 dt dx \leq \int_0^1 \int_0^\infty \lvert\partial_t f(t, x)\rvert^2 \left(1+t^2\right) dt dx < \infty.$$ Hence both terms in the estimate are finite, and we conclude $g\in L^2$. Finally, it’s not hard to show that $$g(x) - f(t, x) = \int_t^\infty \partial_\tau f(\tau, x) d\tau,$$ whose $L^2$ norm necessarily must vanish as $t\to\infty$ due to the convergence of the first integral.

In the previous post, I thought of a possible solution that could utilise a covering lemma. Upon some further research, I found the Vitali Covering Lemma. I will probably revisit the topic of covering theorems later this summer, since they appear to have useful applications (such as this problem). This lemma makes the problem pretty straight-forward, so I’ll omit the solution.

Last time, I was able to reduce the problem to the case where $f$ was a holomorphic mapping of the unit disc to itself with two fixed points. If $z_1$ was a fixed point, one could move it to the origin via the biholomorphic map $w\mapsto\frac{w-z_1}{1-\overline{z_1}w}$, so $f$ fixed both the origin and some other nonzero $z_2\in\mathbb D$. Thus, by the Schwarz lemma, $f$ must be a rotation, and since nontrivial rotations have exactly one fixed point, $f$ is the identity.

I guess I was nearly there, I just forgot about the Schwarz lemma.

Let $V\subset U$ be one of its connected compnonents, and suppose that it’s bounded. We claim that $f(V)=\mathbb D\setminus\lbrace 0\rbrace$. First, $f(V)$ is open in the punctured unit disc by the open mapping theorem. It is also closed in the punctured disc’s relative topology: if $z_0$ is in the interior of the punctured disc and $z_n$ is a sequence in $f(V)$ approaching $z_0$, then we may construct a sequence $w_n$ in $V$ such that $f\left(w_n\right)=z_n$ for all $n$. Since the sequence $w_n$ is bounded, we may assume without loss of generality that it’s convergent by passing to a subsequence. If $w=\lim w_n$, then $f(w)=z_0$. But $w\in\overline V$, and since $V$ is a connected component of $U$ and $U$ is open, $w$ should be in the same connected component as $V$. Hence $w\in V$, and thus $z_0\in f(V)$. We conclude that $f(V)$ is closed and open in the punctured disc, so it’s the entire punctured disc (it’s obviously nonempty).

But we can apply the same argument to points on the boundary of the punctured disc. In particular, consider the sequence $w_n$ in $V$ such that $f\left(w_n\right)=\frac 1{n+1}$ for all $n$. Refining to a convergent subsequence again, setting $w=\lim w_n$ yields $f(w)=0$, contradicting the assumption. We conclude that $V$ could not have been bounded.

The idea is that the only way $f$ can avoid $0$ is if for every sequence $z_n$ such that $f\left(z_n\right)\to 0$, $z_n$ “escapes” to infinity. This can’t happen on a bounded set.