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My function y(x) is $C^{1}$ in $[0,\pi]$ , and $y(0)=y(\pi)=0 $.

I want to show that $\int_{0}^{\pi} |y(x)|^{2} dx \le \int_{0}^{\pi} |y'(x)|^{2} dx $

$Hint:\,Extend \,\, y(x)\, to\, an\, odd\,function $

I used the hint, and hence I wrote $y(x)=\sum_{1}^{\infty} b_{n}sin(nx)$ for $x\in[0,\pi]$.

There is a theorem in my book stating that $\frac{2}{\pi}\int_{0}^\pi|y(x)|^{2} dx = \sum_{1}^\infty |b_{n}|^{2}$. Could I say that the left hand side of the inequality equals to this?

And for the right side I can just differentiate $y(x)$ ? So then $y'(x)=\sum_{1}^\infty b_{n}ncos(nx) $

Can I then say that $\int_{0}^{\pi} |y'(x)|^{2} dx = \sum_{1}^\infty n^{2}|b_{n}|^{2}$.

And hence the inequality should hold, or am I thinking wrong?

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    Check your text again for the Parseval identity. Also, I just had this inequality assigned for homework, and the trick I used was Cauchy-Schwarz. – 2017-02-14
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    So from Parsevals, I should have that: $\int_{0}^\pi |y(x)|^{2} dx = \frac{\pi}{2} \sum_{1}^\infty |b_{n}|^{2} $ ? – 2017-02-14
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    @fejz1234 You did not accept my answer. Is something wrong with it? Is it not useful to you? – 2017-02-22

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Following the hint, extend $y$ to an odd function $\tilde{y}$ in $L^2(\Bbb{T})$ (we can do this because $y(0)=y(\pi)=0$).

Verify (by integration by parts for $n \neq 0$) that $$ \hat{\tilde{y}}(n) = \begin{cases}0 & n=0 \\ \frac{1}{in}\widehat{\tilde{y}'}(n) & n \neq 0\end{cases} $$

Now \begin{align*} \int_{0}^{\pi} |y(x)|^{2} \,dx &= \frac{1}{2} \int_{-\pi}^{\pi} |\tilde{y}(x)|^{2} \,dx \\ &=\pi \| \tilde{y} \|_2^2 \\ &=\pi \sum_{n \in \mathbb{Z}} |\hat{\tilde{y}}(n)|^2 & \text{by Parseval} \\ &=\pi \sum_{\substack{n \in \mathbb{Z} \\ n \neq 0}} \frac{1}{n^2} |\widehat{\tilde{y}'}(n)|^2 \\ &\leq \pi \sum_{n \in \mathbb{Z}} |\widehat{\tilde{y}'}(n)|^2 \\ &=\pi \| \tilde{y}' \|_2^2 & \text{by Parseval} \\ &= \frac{1}{2} \int_{-\pi}^{\pi} |\tilde{y}'(x)|^{2} \,dx \\ &= \int_{0}^{\pi} |y'(x)|^{2} \,dx \end{align*}

Note that the last equality comes from the fact that if $\tilde{y}$ is odd then $\tilde{y}'$ is even.

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    What is $L^2 \(\mathbb{T} \) $ ? I tried to look it up on a list of mathematical symbols, but have not found it. – 2017-02-15
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    @Imago Oh, I don't know if it's a widespread notation, even though that's the one that was used when I took a course dealing with Fourier analysis. Here $\Bbb{T}$ is the unit circle in $\Bbb{C}$ and $L^2(\Bbb{T})$ is just a shorthand for the space of $2\pi$-periodic functions from $\Bbb{R}$ to $\Bbb{C}$ such that $$\int_0^{2\pi}|f(t)|^2\,dt<\infty$$. Similarly we have $L^1(\Bbb{T})$, $C(\Bbb{T})$ and $C^1(\Bbb{T})$. For $f \in L^2(\Bbb{T})$ we define $$\|f\|_2:=\left(\frac{1}{2\pi}\int_0^{2\pi}|f(t)|^2\,dt\right)^{1/2}$$ where the $\frac{1}{2\pi}$ factor is such that $\|1\|_2=1$. – 2017-02-15
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    @Imago We define $\|f\|_1$ and $\|f\|_{\infty}$ in similar usual manners. Since $[0,2\pi]$ is of finite Lebesgue measure, we have $$C^1(\Bbb{T})\subset C(\Bbb{T})\subset L^2(\Bbb{T})\subset L^1(\Bbb{T})$$ The motivation for the notations with $\Bbb{T}$ is that if $f:\Bbb{T}\to\Bbb{C}$ then $\tilde{f}:\Bbb{R}\to\Bbb{C}$, $\tilde{f}(t):=f(e^{it})$ is $2\pi$-periodic, and every $2\pi$-periodic function $\tilde{f}$ is of the form $\tilde{f}(t):=f(e^{it})$ for a $f$ defined on $\Bbb{T}$. We identify $f$ and $\tilde{f}$. – 2017-02-15
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    @Imago (Of course in my first comment you need to add the condition that $f \in L^2(\Bbb{T})$ be measurable.) – 2017-02-15
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    Thank you a lot for you comments! In fact they are very resourceful! – 2017-02-16