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Let $M$ be a subset of $C[0,1]$ such that $f(0)=0$, $f(1)=1$ and $\Vert f\Vert_{\infty}\leq 1$, for every $f \in M$. Prove that $ T(f)=\int_0^1 \! f^2(t) \, \mathrm{d} t $ is a continuous mapping of $M$ into $[0,1]$.

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    Ah yes!do you have any idea of doing this problem?2012-12-27

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It suffices to show that there exist a constant $C>0$ such that for all $f,g \in M$: $|T(f)-T(g)| \leq C \cdot \|f-g\|_\infty$ where $\|\cdot\|_\infty$ denotes the sup norm on $[0,1]$.

We have $|T(f)-T(g)| \leq \int_0^1 \underbrace{|f^2(t)-g^2(t)|}_{|(f(t)+g(t)) \cdot (f(t)-g(t))|} \, dt \leq \|f+g\|_{\infty} \cdot \|f-g\|_{\infty} \cdot \int_0^1 1 \, dt \leq 2 \cdot \|f-g\|_{\infty}$

since $\|f+g\|_{\infty} \leq \|f\|_{\infty}+\|g\|_{\infty} \leq 2$ for all $f,g \in M$.

Remark: Note that

$\int_0^1 \underbrace{f^2(t)}_{\geq 0} \, dt \geq 0 \qquad \qquad \qquad \int_0^1 \underbrace{f^2(t)}_{\leq 1} \, dt \leq 1$

so $T$ is really a mapping of $M$ into $[0,1]$.

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    @Romeo You are welcome. (And anyway, it's good to think about this kind of stufff.)2019-04-07