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Suppose we have a set $S$ which contains all functions $v \in C^{1}[0,1]$ so that $v(0) = 0$ and

$\int_{0}^{1} |v'(x)|^2 dx = 1.$

How can I show that $S$ is bounded with the infinity norm. That is, how can I show that

$ sup_{v\in S}\lVert v\rVert_{\infty} $

exists.

I know that for each $v \in S$ there exists a constant $M(v)$ so that $|v(x)| \le M(v)$ for all $x \in [0,1]$ since $v$ is continuous on a closed interval $[0,1]$. What I fail to see is how $\int_{0}^{1} |v'(x)|^2 dx = 1$ helps me.

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    @macydanim, so is there something very obvious that I might be missing from that integral?2012-12-17

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Let $v\in S$ and $x\in[0,1]$. Then $ |v(x)|=\Bigl|\int_0^xv'(t)\,dt\Bigr|\le\int_0^x|v'(t)|\,dt\le\Bigl(\int_0^x|v'(t)|^2dt\Bigr)^{1/2}\Bigl(\int_0^xdt\Bigr)^{1/2}\le\sqrt x\le1. $ Use has been made of the Cauchy-Schwarz inequality and the fact that $ \int_0^x|v'(t)|^2dt\le\int_0^1|v'(t)|^2dt\le1. $

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    We certainly have $|v(x)|\le\sqrt x$. Can you spot something wrong in the proof? But this does nor mean that $\sqrt x\in S$.2012-12-18
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Another one by contradiction.

Suppose $\sup_{v\in S} || v||_{\infty} =\infty$ That means there exists a sequence $(x_n)\rightarrow \tilde{x}>0$ such that $v(x_n)\rightarrow \infty$. Now the mean value theorem states, that $\exists \xi $ such that $\xi \in [0,{x_n}]$ and $ v'(\xi) = \frac{v(x_n)-v(0)}{{x_n} -0}=\frac{v({x_n})}{{x_n}} $ So we know
$ |v'(\xi)| =|\frac{v({x_n})}{{x_n}}| \rightarrow |\frac{v(\tilde{x})}{\tilde{x}}|= \infty\, \text{ as } n \rightarrow \infty$ Now as $v\in C^1[0,1]$ we know that $v'\in C^0[0,1]$ so $\int_0^1 || v'(x) || dx \rightarrow \infty$. Which is a contradiction.