Let $|\sigma_{n}(x)|\leq K$ $\displaystyle \Rightarrow \frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx\leq 2K^{2}.$ Now, $$\sigma_{n}(x)=\sum_{k=0}^{n}\left(1-\frac{k}{n+1}\right)(a_{k}\cos kx+b_{k}\sin kx).$$ By Parseval's identity, $$\frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx=\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2}).$$ From this it follows that if $m$ is any integer, $m\leq n$ then $$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}\tag{1}$$ Taking $n\to \infty$ and keeping $m$ constant, we get $$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}.\tag{2}$$ I read this thing in some proof of theorem from real analysis book. In this proof i can't understand to derive equation $(1)$ why are they taking that integer $m\leq n$? and from equation $(1)$ how can they get equation $(2)$? Please explain this terminology me! Thanks in advance!
To understand some terminology of proof.
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real-analysis
1 Answers
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If $m \leq n$
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
is not greater than
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
thus since
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx=\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
the inequality follows.
For the second one, note that as $n \to \infty$,
$$\frac{k}{n+1} \to 0$$
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1Can't we directly take $n\to\infty$ to get equation $(1)$ And $(2)$? – 2012-06-18
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0@Kns One does not simply take $n \to \infty$. – 2012-06-18
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0What is a reason? – 2012-06-18
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0It's a joke. I tried to be as clearer as possible in the answer. – 2012-06-18