How could I continue to show the inequality?

Question

Let $g: [0, \pi]\rightarrow \mathbb{R}$ 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.

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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 $$\int_0^{\pi}g^2(x)dx \leq \int_0^{\pi}(g'(x))^2dx$$ ??

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EDIT:

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??

Answer 1

Suppose g is absolutely continuous in $[0, \pi]$ and $g(0) = 0$. Suppose further that $g ‘$ is square integrable. Take the half sine series of g, that is to say we extend g to an odd function G in $[ - \pi , \pi ]$. Then G is periodic, absolutely continuous of period $ 2 \pi $ and the derivative $ G’ $ is Lebesgue integrable and square integrable since $ g ‘$ is square integrable.
The next thing we need to use is the relation between the Fourier coefficients of $G$ and the Fourier coefficients of $ G ‘$.
If (0, b_n ) is the Fourier series of $ G $ , then ( n b_n , 0) is the Fourier coefficients of $ G’ $. More precisely the formal derived series of the Fourier series of G is the Fourier series of $ G ‘$. Now we can invoke Parseval Theorem, since $ G $ and $ G’ $ are square integrable, to get:

$\frac{1}{\pi }\int_{ - \pi }^\pi {{G^2}(x)dx = \sum\limits_{n = 1}^\infty {b_n^2} } $ and

$\frac{1}{\pi }\int_{ - \pi }^\pi {{{(G'(x))}^2}dx} = \sum\limits_{n = 1}^\infty {{n^2}b_n^2} $.

Therefore,

$\frac{2}{\pi }\int_0^\pi {{g^2}(x)dx = \sum\limits_{n = 1}^\infty {b_n^2} } \le \sum\limits_{n = 1}^\infty {{n^2}b_n^2} = \frac{1}{\pi }\int_{ - \pi }^\pi {{{(G')}^2}(x)dx} = \frac{2}{\pi }\int_0^\pi {{{(g')}^2}(x)dx} $

and $\int_0^\pi {{g^2}(x)dx} \le \int_0^\pi {{{(g'(x))}^2}dx} .$

The key to this problem is square integrability and the condition to extend to a continuous periodic odd function. g being absolutely continuous means that its derived function is Lebesgue integrable but may not be square integrable so we make the assumption that $ g' $ is square integrable. If g is smooth then both g and $g'$ are continuous and so are both square integrable. For the extension to odd continuous function we just need to have $g(0) = 0$. We do not need to expand the function as a Fourier series. We do not need to use the convergence of the Fourier series.

Answer 2

Let $G:[-\pi,\pi] \to \mathbb{R}$ be defined by $$ G(x) = \begin{cases} g(x) & \text{if } 0 \le x \le \pi \\ -g(-x) & \text{if } -\pi \le x < 0 \end{cases} $$ You can check that this function is continuous and differentiable (this is where the fact that $g(0) = 0$ is needed). Its derivative is given by $$ G'(x) = \begin{cases} g'(x) & \text{if } 0 \le x \le \pi \\ g'(-x) & \text{if } -\pi \le x < 0 \end{cases} $$ Notice that $G'$ is continuous since $g$ is $C^\infty$.

Because $G$ is odd and $G'$ is even, their Fourier series have the following form : \begin{align} G &\sim \sum_{k=1}^\infty b_k \sin(kx) \\ G' &\sim \sum_{k=1}^\infty A_k \cos(kx) \end{align} Using integration by parts and the fact that $G(\pi) = G(-\pi) = 0$, we can show that $A_k = k b_k$ : \begin{align*} A_k &= \frac{1}{\pi} \int_{-\pi}^\pi G'(x)\cos(kx) \,dx \\ &= \frac{1}{\pi}\left[ G(x)\cos(kx) \right]_{x=-\pi}^{x=\pi} - \frac{1}{\pi} \int_{-\pi}^\pi G(x)(-k\sin(kx)) \,dx \\ &= k\frac{1}{\pi} \int_{-\pi}^\pi G(x)\sin(kx) \,dx \\ &= k b_k \end{align*}

Since $G$ and $G'$ are both continuous and $2\pi$-periodic, we can now use Parseval's formula to conclude the proof : \begin{align*} \int_0^\pi g(x)^2 \,dx &= \frac{1}{2} \int_{-\pi}^\pi G(x)^2 \,dx \\ &= \frac{\pi}{2} \sum_{k=1}^\infty b_k^2 \\ &\le \frac{\pi}{2} \sum_{k=1}^\infty k^2 b_k^2 \\ &= \frac{1}{2} \int_{-\pi}^\pi G'(x)^2 \,dx \\ &= \int_0^\pi g'(x)^2 \,dx \end{align*}