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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??

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  • $\begingroup$ Well, I have the impression that you integral of $g'^2$ could be negative. The reason is that that minus should not be there, since it gets squared with the coefficients. With that, your inequality becomes $\sum(k^2-1)(b_k^2+a_k^2)\geq\frac{a_0^2}{2}$, by carrying the LHS to the right except for the $a_0$ term, putting sums together and dividing both sides by $\frac{\pi}{2}$. Can you show it holds regardless of the coefficients? $\endgroup$ – MickG May 9 '15 at 16:14
  • $\begingroup$ I edited my initial post... I changed the sign... Is it correct now?? @MickG $\endgroup$ – Mary Star May 9 '15 at 16:18
  • $\begingroup$ Yes, I think the post is now correct. As I said, if you carry the LHS sum over to the right and put it together with the one on the RHS and then divide by pi/2, you get an inequality which seems pretty obvious, but can you prove it? I'm not too sure I can… $\endgroup$ – MickG May 9 '15 at 16:21
  • $\begingroup$ @MickG I have the following idea: $\endgroup$ – Mary Star May 9 '15 at 16:50
  • $\begingroup$ $$\sum_{k=1}^{\infty}(k^2-1)(b_k^2+a_k^2)\geq \frac{a_0^2}{2} \\ \Rightarrow \sum_{k=1}^{\infty}(k^2-1)(\frac{4}{\pi^2}\int_0^{\pi}g^2(x)\sin^2 (kx)dx+\frac{4}{\pi^2}\int_0^{\pi}g^2(x)\cos^2 (kx)dx)\geq \frac{\frac{4}{\pi^2}\int_0^{\pi}g^2(x)dx}{2} \\ \Rightarrow \sum_{k=1}^{\infty}(k^2-1)\frac{4}{\pi^2}\int_0^{\pi}g^2(x)(\sin^2(kx)+\cos^2(kx))dx\geq \frac{\frac{4}{\pi^2}\int_0^{\pi}g^2(x)dx}{2}\\ \Rightarrow \sum_{k=1}^{\infty}(k^2-1)\frac{4}{\pi^2}\int_0^{\pi}g^2 (x)dx \geq \frac{\frac{4}{\pi^2}\int_0^{\pi}g^2(x)dx}{2}\\ \Rightarrow \sum_{k=1}^{\infty}2(k^2-1) \geq 0 \text{ which is true.}$$ $\endgroup$ – Mary Star May 9 '15 at 16:50
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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, bn ) is the Fourier series of $ G $ , then ( n bn , 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.

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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*}

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  • $\begingroup$ Just to add that G is continuous in $[- \pi, \pi]$ since $g(0) = g( \pi) $ . Then the differentiated series is automatically the Fourier series of $G '$ . [On account of the Fundamental Theorem of Calculus, the constant term of the Fourier series of G' is 0 bec of $g(0) = g( \pi)$ by integrating $G'$ and use FTC.] You have essentially proved this. $\endgroup$ – user231543 May 10 '15 at 7:59
  • $\begingroup$ This result holds for general $g$ if $g(0) = g( \pi) $, $g$ is absolutely continuous (so we have the Fundamental Theorem of Calculus) and $g'$ is square integrable. [If g is smooth, then it is absolutely continuous and $g'$ is integrable and square integrable . ] The proof is exactly as you have indicated $\endgroup$ – user231543 May 10 '15 at 8:01
  • $\begingroup$ I missed out that g must also satisfied g(0) = 0 so that the extension of g to G is continuous in $[-\pi, \pi]$ $\endgroup$ – user231543 May 10 '15 at 8:21
  • $\begingroup$ I see... Thank you very much!! :-) $\endgroup$ – Mary Star May 10 '15 at 15:57
  • $\begingroup$ 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?? @Math536 $\endgroup$ – Mary Star Jun 15 '15 at 18:45

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