Rudin's Real & Complex - Q9.11 (Fourier)

Question

I have solved most of Question 9.11 of Big Rudin :

Find conditions on $f$ and/or $\widehat{f}$ which ensure the correctness of the following formal argument : If $\varphi(t) ~=~ \frac{1}{2\pi}\int_{-\infty}^\infty f(x) e^{-ixt}\; dx $ and $$ F(x) ~=~ \sum\limits_{k=-\infty}^\infty f(x+2k\pi)$$ then $F$ is periodic, with period $2\pi$, the $n$-th Fourier coefficient of $F$ is $\varphi(n)$, hence $$ F(x) ~=~ \sum\limits_{n=-\infty}^\infty \varphi(n)e^{inx}.$$ In particular, $$\sum\limits_{k=-\infty}^\infty f(2k \pi) ~=~ \sum\limits_{n=-\infty}^\infty \varphi(n).$$ More generally, $$\sum\limits_{k=-\infty}^\infty f(k \beta) ~=~ \alpha \sum\limits_{n=-\infty}^\infty \varphi(n\alpha) ~~~~~\text{if $\alpha > 0, \beta > 0, \alpha \beta = 2\pi$.}$$ What does this last equation say about the limit, as $\alpha \rightarrow 0$, of the right-hand side (for "nice" functions, of course) ? Is this in agreement with the inversion theorem ?

I struggle to see how it meshes with the inversion theorem. I guess I have not quite developed intuition yet. Could anyone share his thoughts about the "What does this last equation say about the limit" bit of the question plz ?

Answer 1

After much thought, I believe I got a solution :

Observe that $\alpha \sum\limits_{n = -\infty}^\infty \varphi(n \alpha)$ is a Riemann sum : $\alpha$ is the lenght of the interval and $\varphi(n\alpha)$ is the evaluation of $\varphi$ on this interval.

If $\alpha \rightarrow 0$, we obtain $\int_{-\infty}^\infty \varphi(x) \; dx$.

Now, as $\alpha \rightarrow 0$ and we must have $\beta \rightarrow \infty$ since $\alpha\beta = 2\pi$. The left-hand side is therefore $$ \sum\limits_{k=-\infty}^\infty f(k \beta) = f(0)$$ since for $\beta$ big enough we have that $f(k \beta) = 0$ for every $k \neq 0$ (here we assume that by 'nice functions' we mean to the very least that $f$ vanishes at infinity).

To sum up, we've got $$ f(0) ~=~ \int_{-\infty}^\infty \varphi(x) \; dx.$$ That is to say $$ f(0) ~=~ \int_{-\infty}^\infty \varphi(x)e^{0} \; dx,$$ which is in agreement with the inversion theorem.

Answer 2

Just to supplement the previous solution, I have similar:

Under the assumption that $f \in S$ where $S$ is the Schwartz space.

Then since $f\in S$ we have that $|f(x)| \le C/|x|^2$ where $x\in\mathbb{R}, x\neq 0$ and $C>0$ is some constant and hence \begin{equation*} \begin{aligned} \lim_{\alpha\to0} \sum_{k=-\infty}^{\infty} f(k\beta) &= \lim_{\alpha\to0} \sum_{k=-\infty}^{\infty} f\left(\frac{k2\pi}{\alpha}\right) \\ &= \lim_{N\to\infty} \sum_{k=-\infty}^{\infty} f\left(2\pi kN\right) \\ &\le \lim_{N\to\infty} \sum_{k=-\infty, k\neq 0}^{\infty} f\left(2\pi kN\right) + \lim_{N\to\infty} f(0) \\ &\le \lim_{N\to\infty} \sum_{k=-\infty, k\neq 0}^{\infty} |f\left(2\pi kN\right)| + f(0) \\ &\le \lim_{N\to\infty} \sum_{k=-\infty, k\neq 0}^{\infty} \frac{C}{|2\pi kN|^2} \\ &= 0 + f(0) = f(0) \quad (1) \end{aligned} \end{equation*} Also, on the other side we have \begin{equation*} \begin{aligned} \lim_{\alpha\to0} \alpha \sum_{n=-\infty}^{\infty} \phi(n\alpha) &\stackrel{(A)}{=} \int_{-\infty}^{\infty} \phi(x) \,dx \\ &= \int_{-\infty}^{\infty} \hat f(x) \,dm(x) \quad \text{ since }\left(\sqrt{2\pi}f(x)\,dm(x) = f(x)\,dx \right) \quad (2) \end{aligned} \end{equation*} Where the (A) is due to the fact that the left side is just a Riemann sum

Hence from (1) and (2) we have that \begin{equation*} \int_{-\infty}^{\infty} \hat f(x) \,dm(x) = f(0) \end{equation*} We have that $f\in S$ hence $f \in L^1$, we can further add the assumption that $\hat f \in L^1$. Hence we are in agreement with the Inversion Theorem.