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here is the question

Find the maximum possible constant so it satisfies the following:

$$\int_0^{2π} (u'(x))^2 dx \geq c\int_0^{2π} (u(x))^2 dx $$

with: $$u\in C^1[0,2π] ,u(0)=u(2π)=0 $$

*My thought was that I should use the Cauchy–Schwarz inequality like this $$(\int_0^{2π} udx)^2 \leq \int_0^{2π}dx\int_0^{2π}u^2dx=2π \int_0^{2π}u^2dx $$

and $$U(x)=\int_0^{y}u'(y)dy$$ so $$(\int_0^{y}u'(y)dy)^2\leq \int_0^{x}dy\int_0^{x}u^2dy=x\int_0^{x}u^2dy$$ , but I am not sure what to do next really, or if I am going anywhere, thanks.

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    Have a look at Wikipedia, *Wirtinger's inequality for functions*. – 2017-02-02
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    found this before but couldn't get much of a help – 2017-02-02
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    The Wikipedia page https://en.wikipedia.org/wiki/Wirtinger's_inequality_for_functions contains the optimal constant ($\frac{1}{4}$) and a proof through Fourier series of the optimal inequality: what else do you need? – 2017-02-02
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    Moreover, it is likely that your inequality is about $C^1$ functions, not just continuous functions. Otherwise, what does $u'(x)$ means? – 2017-02-02
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    Yea, I forgot to write that it belongs to continuous differentiable functions...also figured out the solution myself, thanks anyway. – 2017-02-02

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Note that $\{\frac{1}{\sqrt\pi}\sin(\frac{nx}{2})\}_{n=1}^\infty$ is dense in $A=\{u\in C^1[0,2\pi]: \ u(0)=u(2\pi)=0\}$. Also note \begin{eqnarray} &\int_0^{2\pi}\sin^2(\frac{nx}{2})dx=\int_0^{2\pi}\cos^2(\frac{nx}{2})dx=\pi, \\ &\int_0^{2\pi}\sin(\frac{mx}{2})\sin(\frac{nx}{2})dx=\int_0^{2\pi}\cos(\frac{mx}{2})\cos(\frac{nx}{2})dx=0\text{ for }m\neq n. \end{eqnarray} Let $u\in A$. Then $u$ can be written as $$ u=\sum_{n=1}^\infty\frac{a_n}{\sqrt \pi}\sin(\frac{nx}{2}) $$ and hence $$ u'=\sum_{n=1}^\infty\frac{n}{2\sqrt \pi}\cos(\frac{nx}{2}). $$ where $a_n, n=1,2,3,\cdots$ are reals. So $$ \int_0^{2\pi}u^2dx=\sum_{n=1}^\infty a_n^2, \int_0^{2\pi}(u')^2dx=\sum_{n=1}^\infty \frac{n^2}{4}a_n^2. $$ If $$ \int_0^{2\pi}(u')^2dx\ge C\int_0^{2\pi}u^2dx$$ then one must have $$ \frac{n^2}{4}a_n^2\ge Ca_n^2, n=1,2,\cdots$$ from which one can see that the maximum possible constant for $C$ is $C=\frac14$. It is easy to see that if $u=\frac{1}{\sqrt \pi}\sin(\frac{x}{2})$, the equal sign holds for $C=\frac14$.