# $L^2$ norm inequality

I need some help with this homework question. I was asked to provide an example of a $n$-dimensional subspace $W$ of $L^2[0,1]$ such that all functions in that subspace with $L^2$ norm equal to $1$ satisfy that $\Vert f\Vert_\infty\le\sqrt{n}$. I think I'll have to find an example such that the above is true for all $n$.

I don't know where to start. I'm not sure if this helps, but in previous questions in the homework I showed that if $S$ is a subspace of $C[0,1]$ (which is closed at subspace of $L^2[0,1]$) then $\Vert f\Vert_{\infty}\le M\Vert f\Vert_2$ for all $f\in S$.

In the second part of this question, I will need to show that if $W$ is a $n$-dimensional subspace of $L^2[0,1]$, and all elements of $W$ are continuous functions, then there exists $f\in W$ s.t. $\Vert f\Vert_2=1$ and $\Vert f\Vert_\infty\ge\sqrt{n}$.

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How is it possible that all elements of a subspace to have norm equal to 1? It is not a subspace, it is a subset of a sphere! –  timur Nov 9 '12 at 1:23
Oh no.. sorry I should rephrase –  user9625 Nov 9 '12 at 1:25
Hint for first part: Try to think about the simplest functions you can come up with - say characteristic functions of something. –  user27126 Nov 9 '12 at 1:41
well if $f$ is characteristic function of something then $\Vert f\Vert_\infty\le\sqrt{n}$ will always hold.. I think –  user9625 Nov 9 '12 at 6:13

Hint: for a fixed $n$, take $$W:=\operatorname{Span}\{\chi_{(kn^{-1},(k+1)n^{-1})},0\leqslant k\leqslant n-1\}.$$