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 $[0,1]$.
Topology path-connectedness in $C[0,1]$
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general-topology
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0I tried to put the integration symbol,but I couldn't.So here T(f) is the integration of f^2(t) from 0 to 1. – 2012-12-27
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0thanks Romeo!But norm of f is less than or equal to 1. – 2012-12-27
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1You are welcome. With "The norm of $f$ is less than or equal to 1" do you mean $\Vert f \Vert_{\infty} \le 1$? So we have to consider $C([0,1])$ with the topology given by sup norm, don't we? – 2012-12-27
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0Ah yes!do you have any idea of doing this problem? – 2012-12-27
1 Answers
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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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0Thanks alot Saz!that's a wonderful answer! – 2012-12-28
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0Is it true that $\Vert a \cdot b \Vert_{\infty} \le \Vert a \Vert_{\infty} \cdot \Vert b \Vert_\infty$, where $a,b$ are arbitrary continuous functions? If yes, how can we prove it? Thanks. – 2012-12-30
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1@Romeo Let $x \in [0,1]$ such that $|a(x) \cdot b(x)| = \|a \cdot b\|_{\infty}$ (exists since $a \cdot b$ is continuous and $[0,1]$ compact). Since $|a(x)| \leq \|a\|_{\infty}$, $|b(x)| \leq \|b\|_{\infty}$ we conclude $$\|a \cdot b\|_{\infty} = |a(x) \cdot b(x)| = |a(x)| \cdot |b(x)| \leq \|a\|_{\infty} \cdot \|b\|_{\infty}$$ – 2012-12-30
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0So easy! Thank you very much, I feel idiot. – 2012-12-30
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0@Romeo You are welcome. (And anyway, it's good to think about this kind of stufff.) – 2012-12-30