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Let $R_n(x)=f(x)-S_n(x)$ where $S_n(x)$ is the partial sum of Fourier series of $f(x)$ function.

$\lim_{n \to \infty} <R_n(x),R_n(x)>$ $=0$  $\iff$  $\sum_{n=1}^{\infty}c_i^2 = \int_{-\pi}^{\pi}(f(x))^2dx$ (Parseval identity where $c_i$ s are Fourier coefficients)

It's written that $<R_n(x),R_n(x)>$  $=$ $\sum_{n=1}^{\infty}c_i^2 \cdot\int_{-\pi}^{\pi}(f(x))^2dx$ in my notes. But I cannot identify the reason of this equality.

My main problem is that I cannot see why Parseval identity satisfies $\lim_{n \to \infty} <R_n(x),R_n(x)>$ $=0$ ? Showing in real $L^2[-\pi,\pi]$ is enough for me now. It can be stupid question, I'm sorry. If someone explain it me in easiest way, I will be glad

Thanks

P.S. : I need a proof in the easiest way. Please do not explain with different ways. I’m only an undergraduate student now and my knowledge is really bounded unfortunately. I cannot use anything I didn’t learn in my proofs. Please illuminate me how can I show it via Parseval identity.

user519955
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    Uhm, the equality you don't understand seems to be false, because $\sum_{i=1}^\infty c_i^2\cdot\int_{-\pi}^\pi f^2(x)dx$ is a constant, (i.e. $(\sum_{i=1}^\infty c_i^2)^2$ by Parseval) and doesn't tend to zero... I mean, the theorem is true but I don't get the intermediate statement. – Riccardo Ceccon Oct 29 '18 at 22:25
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    @RiccardoCeccon thank you I couldn’t get that equality in no way. To my surprise, it is false :D – user519955 Oct 29 '18 at 22:50
  • These are things that happen, by the way it's a good way to remember it ;) – Riccardo Ceccon Oct 29 '18 at 23:01
  • @RiccardoCeccon yes you’re absolutely right. Thanks again :) – user519955 Oct 29 '18 at 23:05

1 Answers1

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So, splitting the scalar product we obtain $$\langle R_n(x),R_n(x)\rangle=\langle f(x),f(x)\rangle -2 \langle f(x),S_n(x) \rangle +\langle S_n(x),S_n(x)\rangle.$$ Denoting by $\{e_i\}_{i\in \mathbb{N}}$ the orthonormal basis, this is equal to $$\int_{-\pi}^\pi f^2(x) \ dx +\sum_{i=1}^n c^2_i -2 \sum_{i=1}^n c_i\cdot \langle f(x),e_i\rangle=\int_{-\pi}^\pi f^2(x) \ dx +\sum_{i=1}^n c^2_i -2\sum_{i=1}^n c^2_i =\\ = \int_{-\pi}^\pi f^2(x) \ dx -\sum_{i=1}^n c^2_i,$$ that, by the Parseval Identity tends to zero, because $\sum_{i=1}^nc^2_i$ tends to $\int_{-\pi}^\pi f^2(x) \ dx$ as $n$ tends to $\infty$.

Riccardo Ceccon
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