Let $D = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2\leq 1\}.$ Let for each $n\in\mathbb{Z}, f_n(r,\theta) = r^n e^{in\theta}$. Consider $f(r,\theta) = \sum_{n\in\mathbb{Z}} A_n f_n(r,\theta)$, where $A_n \in \mathbb{C}$. Assume $\sum_{n\in\mathbb{Z}}|A_n| < \infty$ and that $\lim\limits_{r\to 1} f(r,\theta) =: g(\theta)$ uniformly in $\theta$. Prove that $f$ satisfies $f_{rr} + \frac{1}{r} f_r + \frac{1}{r^2} f_{\theta\theta} = 0$ and that $A_n = \frac{1}{2\pi} \int_{-\pi}^\pi g(\theta) e^{-in\theta}d\theta.$

Edit: One should actually have $f_n(r,\theta) = r^{|n|}e^{in\theta}$ and the limit should be $\lim\limits_{r\to 1^-} f(r,\theta) =: g(\theta)$.

One can show the first result by differentiating f. For instance, $f_r(\theta) = \sum_{n\in \mathbb{Z}} nr^{n-1}e^{in\theta}, f_{rr})\theta) = \sum_{n\in\mathbb{Z}}n(n-1)r^{n-2}e^{in\theta},$ and $f_{\theta\theta} = \sum_{n\in\mathbb{Z}} -n^2 r^ne^{in\theta}$ . From this, it's clear that $f_{rr} + \frac{1}r f_r + \frac{1}{r^2} f_{\theta\theta} = 0$.

Is there a more formal reason why I can just take the derivative in the infinite sum? If this doesn't always hold, then why might it not hold?

For the second result, I'm not sure how to show it. I'm not sure what $\lim\limits_{r\to 1} u(r,\theta) = g(\theta)$ uniformly in $\theta$ means. Perhaps there's a typo and it means $\lim\limits_{r\to 1} f_n(r,\theta) = g(\theta)$ uniformly in $\theta$? Or am I misunderstanding something?

I know that $\int_{-\pi}^\pi e^{in\theta} e^{-im\theta} = \begin{cases}0, &\text{ if $n\neq m$}\\ 2\pi, &\text{if $n=m$}\end{cases}$. So the claim would follow easily if I could just show that $\begin{align}\int_{-\pi}^\pi g(\theta)e^{-in\theta} d\theta &=\int_{-\pi}^\pi\lim\limits_{r\to 1} f(r,\theta) e^{-in\theta} d\theta \\ &= \int_{-\pi}^\pi \lim\limits_{r\to 1} \sum_{m\in\mathbb{Z}}A_m f_m(r,\theta) e^{-in\theta}d\theta\\ & = \int_{-\pi}^\pi \sum_{m\in\mathbb{Z}}\lim\limits_{r\to 1} A_m f_m(r,\theta)e^{-in\theta}d\theta\\ & = \int_{-\pi}^\pi \sum_{m\in\mathbb{Z}} A_m e^{i(m-n)\theta}d\theta\\ &= \sum_{m\in\mathbb{Z}}\int_{-\pi}^\pi A_m e^{i(m-n)\theta}d\theta = 2\pi A_n.\end{align}$

My issue is a lot of these equalities seem unjustified, and it shouldn't be necessary to use theorems like the dominated convergence theorem to justify them. For instance, the ones I'd like to justify more formally are the third and fifth equalities.

For the third equality, I tried proving this using the Weierstrass M-test, but I can't seem to uniformly bound the $f_n(r,\theta)$'s, regardless of whether $r > 1$ or $r < 1$.

For the fifth equality, we prove that if $f_n$ and $f$ are continuous functions and $f_n\to f$ uniformly, then $g = \lim\limits_{n\to\infty} g_n$, where $g_n(x) = \int_a^x f_n(t)dt, g(x) = \int_a^x f(t)dt.$ Let $\epsilon > 0$. Since $f_n\to f$ uniformly, we may find $N$ so that $n\ge N\Rightarrow \lVert f_n - f\rVert_\infty < \epsilon/(b-a+1)$. Then $|g_n(x) - g(x)| = |\int_a^x f_n(t) - f(t)dt|\leq \int_a^x |f_n(t)-f(t)|dt \leq \int_a^x \lVert f_n - f\rVert_\infty dt \leq \int_a^b \lVert f_n - f\rVert_\infty dt < \epsilon,$ so $\lVert g_n - g\rVert_\infty \leq \epsilon$, proving that $g_n\to g$ uniformly.

And by the Weierstrass M-test, since for all $m\in\mathbb{Z}, |A_m e^{i(m-n)\theta}| \leq |A_m|$ and $\sum_{m\in\mathbb{Z}} |A_m| < \infty$, it follows that $\sum_{|m|\leq N} A_m e^{i(m-n)\theta}\to \sum_{m\in \mathbb{Z}} A_me^{i(m-n)\theta}$ uniformly as $N\to \infty$.

  • $\begingroup$ This looks like a complex-analysis question in disguise. With $z=r^{i \theta}$ you have $f(z) = \sum A_n z^n$, which is holomorphic. $f_{rr} + \frac{1}{r} f_r + \frac{1}{r^2} f_{\theta\theta} = 0$ is the Laplace equation in polar coordinates, and $A_n = \frac{1}{2\pi} \int_{-\pi}^\pi g(\theta) e^{-in\theta}d\theta$ is Cauchy's integral formula for the derivatives. $\endgroup$
    – Martin R
    May 16 at 8:12
  • $\begingroup$ @MartinR thanks. Could you verify whether the solution I obtained for the following question is correct (you don't have to provide any hints as to formally justify the last part): math.stackexchange.com/questions/4449990/…? $\endgroup$
    – user3472
    2 days ago
  • $\begingroup$ Do you really mean $\sum_{n\in\mathbb{Z}}$ or should it be $\sum_{n\in\mathbb{N}}$? If you allow negative indices then $f$ is not defined for $z=0$. $\endgroup$
    – Martin R
    2 days ago
  • $\begingroup$ @MartinR could you elaborate? Which question are you referring to? Is it the current one? $\endgroup$
    – user3472
    2 days ago
  • $\begingroup$ I am referring to this question. For negative $n$ and $r=0$ is $r^n e^{in\theta}$ not defined. $\endgroup$
    – Martin R
    2 days ago


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.