upper bound for Fourier series

Question

I have a question concerning the upper bound of the sum of the Fourier series and how to prove the following.

I have the sum as follows:

$$\int^\pi_{-\pi} |\frac{1}{n} \sum_{k = m-n}^{m-1} S_k|^2 \leq \int_{-\pi}^{\pi} |f|^2$$

where $ f \in L^2[-\pi, \pi]$ with $m,n \in \mathbb{N} , m>n$, and

$$S_p = \sum_{k = -p}^{p} \hat f_k e^{ikx}$$

where $\hat{f}_k$ is the $k$'th Fourier series coefficient defined as

$$\hat f_k = \frac{1}{2\pi}\int^{\pi}_{-\pi} f(y) e^{-iky} dy$$

I've also tried using the fact that $e^{ikx}$ forms a basis in $L^2$ but I haven't achieved much

edit: changed $c_k$ to $\hat f$

Answer 1

First note that $$ \int_{-\pi}^\pi |S_k|^2\,dx=2\pi\sum_{j=-k}^{k}|\hat{f}_k|^2\le 2\pi\sum_{j=-\infty}^{\infty}|\hat{f}_k|^2\le \int_{-\pi}^\pi |f|^2\,dx, $$ Then, using the fact that $$ |a_1+\cdots+a_n|^2\le n(|a_1|^2+\cdots+|a_n|^2) $$ we obtain $$ \frac{1}{n^2}\int_{-\pi}^\pi \left|\sum_{k=m-n}^{m-1}S_k\,\right|^2\,dx \le \frac{1}{n}\sum_{k=m-n}^{m-1}\int_{-\pi}^\pi |S_k|^2\,dx\le \int_{-\pi}^\pi |f|^2\,dx. $$

Proof of the Fact: $$ |a_1+\cdots+a_n|^2\le (|a_1|+\cdots+|a_n|)^2=\sum_{i,j=1}^n|a_i||a_j|\\ \le \frac{1}{2}\sum_{i,j=1}^n(|a_i|^2+|a_j^2|)= n(|a_1|^2+\cdots+|a_n|^2) $$ $$ $$

Answer 2

Let $$A_j = \sum_{p=0}^{j-1}S_p,$$ the sum of the first $j$ partial sums of the Fourier series of $f$. Observe that $$\sum_{p=m-n}^{m-1}S_p = \sum_{p=0}^{m-1}S_p - \sum_{p=0}^{m-n-1}S_p = A_m - A_{m-n}$$ Let's find a simpler formula for $A_j$: $$\begin{aligned} A_j &= \sum_{p=0}^{j-1}S_p \\ &= \sum_{p=0}^{j-1} \sum_{k=-p}^{p} \hat{f}_k e^{ikx} \\ &= \sum_{k=-(j-1)}^{j-1}(j-|k|)\hat{f}_k e^{ikx} \end{aligned}$$ The last line is obtained by reversing the order of summation and counting how many times the term $\hat{f}_k e^{ikx}$ occurs for each $k \in \{-(j-1), \ldots, -1, 0, 1, \ldots, j-1\}$. Drawing a picture helps if this isn't clear.

Now let's simplify $A_{m} - A_{m-n}$: $$\begin{aligned} A_{m} - A_{m-n} &= \sum_{k=-(m-1)}^{m-1}(m-|k|)\hat{f}_k e^{ikx} - \sum_{k=-(m-n-1)}^{m-n-1}(m-n-|k|)\hat{f}_k e^{ikx} \\ &= \sum_{k=-(m-1)}^{m-1} a_k \hat{f}_k e^{ikx} \end{aligned}$$ where $$a_k = \begin{cases} m-|k| & \text{if }m-n \leq |k| \leq m-1 \\ n & \text{if }|k| \leq m-n-1 \\ \end{cases}$$ Now let's look at the expression inside the absolute values in your integral. Let's call it $g(x)$: $$\begin{aligned} g(x) &= \frac{1}{n}\sum_{k=m-n}^{m-1}S_k \\ &= \frac{1}{n}(A_{m} - A_{m-n}) \\ &= \frac{1}{n}\sum_{k=-(m-1)}^{m-1} a_k \hat{f}_k e^{ikx} \\ &= \sum_{k=-(m-1)}^{m-1} b_k \hat{f}_k e^{ikx} \\ \end{aligned}$$ where $$b_k = \frac{a_k}{n} = \begin{cases} \displaystyle\frac{m-|k|}{n} & \text{if }m-n \leq |k| \leq m-1 \\ 1 & \text{if }|k| \leq m-n-1 \\ \end{cases}$$ Note that $0 \leq \displaystyle \frac{m-|k|}{n} \leq 1$ when $m-n \leq |k| \leq m-1$. Therefore $|b_k| \leq 1$ for all $|k| \leq m-1$. Consequently, $$\sum_{k=-(m-1)}^{m-1}|b_k|^2 |\hat{f}_k|^2 \leq \sum_{k=-(m-1)}^{m-1} |\hat{f}_k|^2 \leq \sum_{k=-\infty}^{\infty} |\hat{f}_k|^2$$

All that remains is to apply Parseval's theorem to both sides of the inequality: $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|g(x)|^2\ dx = \sum_{k=-(m-1)}^{m-1}|b_k|^2 |\hat{f}_k|^2 \leq \sum_{k=-\infty}^{\infty} |\hat{f}_k|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\ dx$$ Multiplying by $2\pi$ gives the desired result.