- #1
mathmari
Gold Member
MHB
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Hey! 
Let $f: [0, \pi]\rightarrow \mathbb{R}$ a $C^{\infty}$ function for which the following stands:
$$g(0)=0 \ \ , \ \ g(\pi)=0$$
I have to show that $$\int_0^{\pi}g^2(x)dx \leq \int_0^{\pi}(g'(x))^2dx$$ using Parseval's formula.
I have done the following:
The Fourier series of $g$ is $$g \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}\left (a_k \cos (kx)+b_k \sin (kx)\right )$$ where $$a_0=\frac{2}{\pi}\int_0^{\pi}g(x)dx \\ a_k=\frac{2}{\pi}\int_0^{\pi}g(x)\cos (kx)dx \ \ , \ \ k=1, 2, \dots \\ b_k=\frac{2}{\pi}\int_0^{\pi}g(x)\sin (kx)dx \ \ , \ \ k=1, 2, \dots $$
From Parseval's formula we have the following:
$$\int_0^{\pi}\left (\frac{a_0}{2}\right )^2dx+\sum_{k=1}^{\infty}\left (a_k^2\int_0^{\pi}\cos^2 (kx)dx+b_k^2\int_0^{\pi}\sin^2 (kx)dx\right )=\int_0^{\pi}g^2(x)dx \\ \Rightarrow \int_0^{\pi}g^2(x)dx=\frac{a_0^2}{4}\pi+\frac{\pi}{2}\sum_{k=1}^{\infty}\left (a_k^2+b_k^2\right )$$
The Fourier series of $g'$ is $$g'\sim \sum_{k=1}^{\infty}\left (kb_k\cos (kx)-ka_k\sin (kx)\right )$$
From Parseval's formula we have the following:
$$\sum_{k=1}^{\infty}\left (k^2b_k^2\int_0^{\pi}\cos^2 (kx)dx+(-k)^2a_k^2\int_0^{\pi}\sin^2 (kx)dx\right )=\int_0^{\pi}(g'(x))^2dx \\ \Rightarrow \int_0^{\pi}(g'(x))^2dx=\frac{\pi}{2}\sum_{k=1}^{\infty}\left (k^2b_k^2+k^2a_k^2\right )$$
Is this correct?? Or have I done something wrong at the application of Parseval's formula??
How could I continue to show the inequality $$\int_0^{\pi}g^2(x)dx \leq \int_0^{\pi}(g'(x))^2dx$$ ?? (Wondering)
Let $f: [0, \pi]\rightarrow \mathbb{R}$ a $C^{\infty}$ function for which the following stands:
$$g(0)=0 \ \ , \ \ g(\pi)=0$$
I have to show that $$\int_0^{\pi}g^2(x)dx \leq \int_0^{\pi}(g'(x))^2dx$$ using Parseval's formula.
I have done the following:
The Fourier series of $g$ is $$g \sim \frac{a_0}{2}+\sum_{k=1}^{\infty}\left (a_k \cos (kx)+b_k \sin (kx)\right )$$ where $$a_0=\frac{2}{\pi}\int_0^{\pi}g(x)dx \\ a_k=\frac{2}{\pi}\int_0^{\pi}g(x)\cos (kx)dx \ \ , \ \ k=1, 2, \dots \\ b_k=\frac{2}{\pi}\int_0^{\pi}g(x)\sin (kx)dx \ \ , \ \ k=1, 2, \dots $$
From Parseval's formula we have the following:
$$\int_0^{\pi}\left (\frac{a_0}{2}\right )^2dx+\sum_{k=1}^{\infty}\left (a_k^2\int_0^{\pi}\cos^2 (kx)dx+b_k^2\int_0^{\pi}\sin^2 (kx)dx\right )=\int_0^{\pi}g^2(x)dx \\ \Rightarrow \int_0^{\pi}g^2(x)dx=\frac{a_0^2}{4}\pi+\frac{\pi}{2}\sum_{k=1}^{\infty}\left (a_k^2+b_k^2\right )$$
The Fourier series of $g'$ is $$g'\sim \sum_{k=1}^{\infty}\left (kb_k\cos (kx)-ka_k\sin (kx)\right )$$
From Parseval's formula we have the following:
$$\sum_{k=1}^{\infty}\left (k^2b_k^2\int_0^{\pi}\cos^2 (kx)dx+(-k)^2a_k^2\int_0^{\pi}\sin^2 (kx)dx\right )=\int_0^{\pi}(g'(x))^2dx \\ \Rightarrow \int_0^{\pi}(g'(x))^2dx=\frac{\pi}{2}\sum_{k=1}^{\infty}\left (k^2b_k^2+k^2a_k^2\right )$$
Is this correct?? Or have I done something wrong at the application of Parseval's formula??
How could I continue to show the inequality $$\int_0^{\pi}g^2(x)dx \leq \int_0^{\pi}(g'(x))^2dx$$ ?? (Wondering)
