Prove: $\lim_{n\to \infty}\|*_{i=1}^nf(x)\|_2=0$

Question

The following is a part of a question which I have answered

Let $ f(x)= \begin{cases} \frac{1}{2}, \mid x \mid<\frac{1}{2} \\ 0, \mid x \mid\geq \frac{1}{2} \end{cases} $

1. Find Fourier transform:

   For $\omega=0: \hat{f}(0)=\frac{1}{4\pi}$

   For $\omega\neq 0:\hat{f}(\omega) =\frac{\sin(\frac{\omega}{2})}{\omega}$

2. Find the Fourier transform of the convolution $f_n(x) = *_{i=1}^{n}f(x)=f(x)*f(x)*\cdots*f(x)$

   For $\omega=0: \hat{f_n}(0)=\frac{2\pi }{(4\pi)^n}$

   For $\omega\neq 0:\hat{f_n}(\omega) = \frac{\sin^n(\frac{\omega}{2})}{\omega^n}$

Now I am asked to prove that $\lim_{n\to \infty}\|*_{i=1}^nf(x)\|_2=0$, I know that $\lim_{\omega\to \infty}F(\omega)=0$ how should I approach this?

Answer 1

This answer won't be the most helpful to the OP (apologies), but it may be helpful for others, so I'll post it anyways.

Let $f_n(x) = *_{i=1}^n f(x)$. The Fourier transform is an isometry from $L^2$ to $L^2$. Hence, if we want to compute $\lim_{n\rightarrow\infty} \|f_n\|_2$, it suffices to compute $\lim_{n\rightarrow\infty} \|\hat{f_n}\|_2$.

Since $\hat{f_n}(\omega) = \frac{\sin^n(\omega/2)}{\omega^n}$, we want to compute $$\lim_{n\rightarrow\infty} \|\hat{f_n}\|_2^2 = \lim_{n\rightarrow\infty} \int_\mathbb{R} |\hat{f_n}(\omega)|^2 \,d\omega$$

The goal, then, is to bound $|\hat{f_n}(\omega)|^2$ by some $g(\omega) \in L^1(\mathbb{R})$.

Since we have for all $x \in \mathbb{R}$, $|\sin(x)| \leq |x|$, we note that $|\hat{f_n}(\omega)|^2 \leq \left|\frac{(\omega/2)^{2n}}{\omega^{2n}}\right| \leq 1$.

Now, we also have the bound $|\hat{f_n}(\omega)|^2 \leq \frac{1}{\omega^{2n}} \leq \frac{1}{\omega^2}$.

Applying both of these bounds, we see that if we let $$g(\omega) = \begin{cases} 1 & \omega \in [-1,1] \\ \frac{1}{\omega^2} & \omega \notin [-1,1]\end{cases}$$ then $g \in L^1$ and $|\hat{f_n}(\omega)|^2 \leq g(\omega)$ for all $\omega \in \mathbb{R}$. Then, apply the dominated convergence theorem to find that

$$\lim_{n\rightarrow\infty} \|f_n\|_2^2 = \lim_{n\rightarrow\infty} \|\hat{f_n}\|_2^2 = \lim_{n\rightarrow\infty} \int_\mathbb{R} |\hat{f_n}(\omega)|^2 \,d\omega = \int_\mathbb{R} \lim_{n\rightarrow\infty} |\hat{f_n}(\omega)|^2 \,d\omega = 0$$