What would be a counterexample that one could use to disprove the statement that $\pi: C^1[a,b] \rightarrow C[a,b]$ given by $\pi(f) = f'$ is continuous? The metric is the usual sup norm one.
Disproving continuity for a map in $C[a,b]$
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real-analysis
functional-analysis
2 Answers
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HINT: Consider the functions $f_n(x)=\frac1n\sin nx$ for $n\in\Bbb Z^+$ on the interval $[0,1]$. They converge in the sup norm to a very nice function $f$; what is that function? Does the sequence $\langle f_n'(x):n\in\Bbb Z^+\rangle$ converge to $f\,'$?
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Hint:
Take a look at the sequence $f_n=\sin(nt)$.
The sequence $\{f_n:n\in\mathbb{N}\}$ is bounded in $(C^1([a,b]),\Vert\cdot\Vert_\infty)$,
The sequence $\{\pi(f_n):n\in\mathbb{N}\}$ is not bounded in $(C([a,b]),\Vert\cdot\Vert_\infty)$.
Continuous linear operators maps bounded sets to bounded ones.
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0Some details would be helpful – 2012-09-19
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0@BrianM.Scott: The functions $f_{n}$ oscillates between 1 and -1 quicker as $n$ goes large, if you use $\sup$ norm then the desired function would be infinitely close to any value in $[-1,1]$ for any $x\in [a,b]$. I do not see how this is possible. – 2012-09-19
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0@user32240: I misread it as $1/n$ for the behavior of the function but read it right for the derivative. Ignore what I said. – 2012-09-19
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0@BrianM.Scott: This is helpful; I tried to construct a proof but was hindered by precisely this kind of behavior. Thanks. – 2012-09-19