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I'm reading section 4.26 in Big Rudin, but I have two questions.
Suppose $f$ is in $L^1(T)$. This means $f$ is the class of all complex, $2\pi$-periodic, and Lebesgue measurable functions on $R^1$ for which the norm $$||f||_p=\{\frac{1}{2\pi}\int_{-\pi}^{\pi} |f(t)|^p dt\}^{1/p}$$ is finite.
For any $f\in L^1(T)$, Fourier coefficients of $f$ is define by the formula $$\hat{f(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)e^{-int} dt$$ where $n$ is an integer. The Fourier series of $f$ is $$\sum_{-\infty}^{\infty} \hat{f(n)}e^{int},$$ and its partial sums are $$s_N=\sum_{-N}^{N} \hat{f(n)}e^{int}$$ where $N$ is a natural number (include 0).
It is the fact (the Parseval theorem) that for any $f, g \in L^2(T)$, $$ \sum_{n=-\infty}^{\infty} \hat{f(n)}\overline{\hat{g(n)}}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)\overline{g(t)}dt$$ holds.
Next, Rudin says $$\lim_{N \to \infty}||f-s_N||_2=0$$ since a special case of the Parseval theorem yields $${||f-s_N||_2}^2=\sum_{|n|>N}|\hat{f(n)}|^2$$.

I have questions:
1) how did Rudin derive the last identity from the Parseval theorem? Is this equation obtained by putting $f(t)=g(t)=|f-s_N|$? (but I can't reach the answer)
2) Why the last equation gives the limit? I see $||f-s_N||$ decreases if taking the limit, but I don't know what it means to converge to zero.

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1 Answers1

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Note $$\hat{s}_N(m) = \sum_{\lvert n\rvert \le N} \hat{f}(n)\cdot\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(n-m)t}\, dt = \sum_{\lvert n\rvert \le N} \hat{f}(n)\delta_{nm}$$ where $\delta_{nm}$ equals $1$ when $n = m$ and equals $0$ otherwise. If $\lvert m\rvert > N$, then $\delta_{nm} = 0$ for every $n$ with $\lvert n \vert \le N$. Therefore, $\hat{s}_N(m) = 0$ whenever $\lvert m\rvert > N$. On the other hand, if $\lvert m\rvert \le N$, then $\sum_{\lvert n \rvert \le N} \hat{f}(n)\delta_{nm}$ reduces to $\hat{f}(m)$. It follows that $\hat{f}(m) - \hat{s}_N(m)$ is $\hat{f}(m)$ for $\lvert m\rvert > N$ and $0$ for $\lvert m \rvert \le N$. By Parseval's theorem, $$\|f - s_N\|_2^2 = \sum_{n\, =\, -\infty}^\infty \lvert\hat{f}(n) - \hat{s}_N(n)\rvert^2 = \sum_{\lvert n\rvert > N} \lvert \hat{f}(n)\rvert^2$$ Since the sequence $(\hat{f}(n))_{n = 1}^\infty$ is square summable, given a positive number $\varepsilon$, there corresponds a $k$ such that for all $N > k$, $\sum_{\lvert n \rvert > N} \lvert \hat{f}(n)\rvert^2 < \varepsilon$. Thus $$\lim_{N\to \infty} \sum_{\lvert n \rvert > N} \lvert \hat{f}(n)\rvert^2 = 0$$ implying the $\lim_{N\to \infty}\|f - s_N\|_2 = 0$.

kobe
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