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青青子衿
Posted at 2020-6-10 15:09:03
回复 3# APPSYZY
将\(f(x)\)在区间\([0,1]\)上展开为正弦级数
\[f(x)=\sum_{n=1}^{\infty}b_n\sin\left(n\pi\,\!x\right)\]
则有
\[f'(x)=\sum_{n=1}^{\infty}\big(n\pi\big)\,\!b_n\sin\left(n\pi\,\!x\right)\]
由帕塞尔(Parseval)等式，可知
\begin{align*}
\int_0^1\!f^2(x)\,\mathrm{d}x&=\dfrac{1}{2}\sum_{n=1}^{\infty}{b_n}\!^2\\
\int_0^1\!\Big(f'(x)\Big)^2\,\mathrm{d}x&=\dfrac{1}{2}\sum_{n=1}^{\infty}\big(n\pi\big)^2{b_n}\!^2=\dfrac{\pi^2}{2}\sum_{n=1}^{\infty}n^2{b_n}\!^2\\
&\geqslant\dfrac{\pi^2}{2}\sum_{n=1}^{\infty}{b_n}\!^2=\pi^2\cdot\dfrac{1}{2}\sum_{n=1}^{\infty}{b_n}\!^2\\
&=\pi^2\int_0^1\!f^2(x)\,\mathrm{d}x
\end{align*}
于是，有\(\,\displaystyle\int_0^1\!f^2(x)\,\mathrm{d}x\leqslant \dfrac{1}{\pi^2}\int_0^1\!\Big(f'(x)\Big)^2\,\mathrm{d}x\,\)，
等号当且仅当\(\,f(x)=\sin(\pi\,x)\,\)时取得. |
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abababa
Posted at 2020-6-12 12:04:43
