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Let $f: [0,1] \to \mathbb{R}$ be continuously differentiable with $f(0)=0$. Prove that $$\Vert f \Vert^{2} = \int_{0}^{1} (f'(x))^{2}dx$$

Here $\Vert f \Vert$ is given by $\sup\{|f(t)|: t \in [0,1]\}$.

I see how to prove that $(\int_{0}^{1} (f'(x))^{2}dx)^{1/2}$ is an upper bound for $f(t)$, where $t \in [0,1]$, from this question.

However, I am not sure how to prove this is the least upper bound.

cm007
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  • How about $f(x)=\sin{2n\pi x}$ with large $\alpha$? The LHS is $1$ but the RHS is (I think) 2n^2\pi^2$. – Aphelli Jun 27 '20 at 18:22

1 Answers1

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You cannot prove it, since it is false. Take $f\colon[0,1]\longrightarrow\Bbb R$ defined by $f(x)=x(1-x)$. Then $\|f\|=\frac14$, but$$\int_0^1(f'(x))^2\,\mathrm dx=\int_0^1(1-2x)^2\,\mathrm dx=\frac13.$$

José Carlos Santos
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