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I am trying to show that $$ \lim \limits_{k \rightarrow \infty } \int \limits_{-\pi}^\pi f(x)\cos(kx)dx = 0 \textit{ and } \lim \limits_{k \rightarrow \infty } \int \limits_{-\pi}^\pi f(x)\sin(kx)dx = 0 $$ I am assuming that $f \in L^2(-\pi, \pi)$ and that the sequence $$ \lbrace \phi_k\rbrace_{k=1}^\infty= \lbrace \frac{1}{\sqrt{2 \pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac{\sin(x)}{\sqrt{\pi}},..., \frac{\cos(kx)}{\sqrt{\pi}}, \frac{\sin(kx)}{\sqrt{\pi}},... \rbrace $$ is orthonormal and in $L^2(-\pi, \pi)$. I have started the problem by using the Bessel inequality, which says that $\sum \limits _ {k=1}^\infty h_k^2 \leq \parallel f \parallel_2^2 $ where $h_k = \langle f, \phi_k \rangle$. I know that all elements of the sequence are orthogonal so $\langle f_k, f_l \rangle= 0$ where $k,l \in R$ and $k \neq l$. I also now that the sequence is orthonormal so $\parallel f \parallel_2 = 1$.

I can rewrite the Bessel inequality as follows $$ \sum \limits _ {k=1}^\infty h_k^2 \leq \parallel f \parallel_2^2 \Rightarrow \sum \limits _ {k=1}^\infty \langle f, \phi_k \rangle^2 \leq 1 $$ This can be simplified further to $$ \sum \limits _ {k=1}^\infty \int \limits_{-\pi}^\pi (f \phi_k ) \int \limits_{-\pi}^\pi (f \phi_k )\leq 1 \Rightarrow \sum \limits _ {k=1}^\infty \int \limits_{-\pi}^\pi (f \phi_k )^2 \leq 1 $$ From here I am stuck. Was my methodology correct?

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For notation, I'll denote the orthonormal sequence by $\{e_k\}_{k\in\mathbb{Z}}$ where $e_k = \frac{e^{ik}}{\sqrt{\pi}}$. This is clearly an orthonormal sequence. By Besell's inequality, we have that $\sum_{k=1}^{\infty} (f,e_k)^2 \le ||f||_2^2$. Since $f \in L^2$, $||f||_2 < \infty$, and so Bessell's inequality tells us in particular that this sum converges. Hence $\lim_{k\to \infty} (f,e_k)^2 = 0$, from which we may conclude that $(f,e_k) \to 0$. Use this together with the definition of $(f,e_k)$ to conclude that the integrals go to $0$.

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I understand what you are saying for the most part, but how can you say that the $\lim (f,e_k) = 0$ if the sum converges. Would we need to know what the sum converges to before we can say what the limit converges to? –  rioneye Nov 28 '12 at 2:48
    
The only other theorem I have says that if the sequence converges then $\lim \limits_{k \rightarrow \infty} \parallel f_k - f\parallel = 0$. Is that what you are using? –  rioneye Nov 28 '12 at 3:06
    
If $\{c_n\}$ is a sequence of complex numbers such that $\sum_{n=1}^{\infty} c_n$ converges, then $\lim_{n \to \infty} c_n = 0$. –  anonymous Nov 28 '12 at 3:08
    
I don't think the sequence I provided had any complex values, so that theorem wouldn't hold, would it? –  rioneye Nov 28 '12 at 3:11
    
$(f,e_k)$ is just some number. It's obtained by integrating f against e^-ikx. –  anonymous Nov 28 '12 at 3:16
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