Prove Inequality w/ Parseval's Formula

[mathmari]

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)

 

[GJA]

Hi mathmari!

We want to use the fact that $$g$$ satisfies the boundary conditions:

$$g(0)=g(\pi)=0$$

The way to do this is to note that this b.c. implies $$g$$ can be expressed as a sine series (i.e. we don't need to use any cosine terms). For a reference on this see the Example section in

Sturm–Liouville theory - Wikipedia, the free encyclopedia

Let me know if anything is unclear/not quite right.

 

[mathmari]

Hi mathmari!

We want to use the fact that $$g$$ satisfies the boundary conditions:

$$g(0)=g(\pi)=0$$

The way to do this is to note that this b.c. implies $$g$$ can be expressed as a sine series (i.e. we don't need to use any cosine terms). For a reference on this see the Example section in

Sturm–Liouville theory - Wikipedia, the free encyclopedia

Let me know if anything is unclear/not quite right.

Ok... Thank you! (Smile)

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When we have to show with the Parseval's formula an inequality in which case do we have to take an expansion of the function that is involved at the inequality?? (Wondering)