# proving $L^\infty$ norm inequality (disprove $\Vert f\Vert_\infty\le\sqrt{n}$)

There are three parts in this question, I've done the first two but not sure about the third one. Also see $L^2$ norm inequality.

In the third part, I am asked 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 $\Vert f\Vert_\infty\ge\sqrt{n}$.

The hint says use the result from part 2. In part 2 I showed that there exists $f\in W$ such that $\Vert f\Vert_2=1$ and $\Vert f\Vert_\infty=\sup_{x\in[0,1]}\vert F(x)\vert$ (since $f$ continuous, this is just $\max(\cdot)$). Note that $(f_1,\dots,f_n)$ is orthonormal basis of $W$ w.r.t the $L^2$ inner product. We regard this as the map $F:[0,1]\mapsto\mathbb{C}^n$. $\vert F\vert$ is the Euclidean norm of $F(x)$ in $\mathbb{C}^n$.

Thanks for any suggestions.

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By your definition, $|F(x)|^2=\sum_{i=1}^n|f_i(x)|^2$. Then $\int_0^1|F(x)|^2 dx=\sum_{i=1}^n\int_0^1|f_i(x)|^2 dx=n$. It implies that $\sup_{x\in[0,1]}|F(x)|\ge \sqrt{n}$.