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Let $\mathbb 1: (C^1([0,1]), \|f\|:=\|f\|_\infty + \|f'\|_\infty)\to (C^1([0,1]),\|\cdot\|_\infty)$ denote the identity mapping between $C^1([0,1])$ with different norms. Then $f$ is linear, continuous and one-to-one, but the inverse Operator $\mathbb 1^{-1}$ is not continuous.

I am trying to convince myself that the inverse operator is indeed not continuous, but I don't know how. I tried to show that it is not bounded, but I didn't really know how to proceed after writing down the definition of the operator norm for $\mathbb 1^{-1}$.

How can I show that $\mathbb 1^{-1}$ is not continuous? Thanks.

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To show the inverse operator is unbounded, you need to find functions $f_n \in C^1([0,1])$ such that $\left\| f_n \right\|_\infty \le 1$, but $\left\| f_n \right\|_\infty + \left\| f_n' \right\|_\infty \ge n$.

Try letting $f_n(x) = x^n$.

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  • $\begingroup$ This is a good idea for an example, thanks! $\endgroup$ – dinosaur May 23 '14 at 20:27
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It's not bounded because you can find a family of $C^1$ functions on $[0,1]$ which, say, take values in $[0,1]$ but whose derivatives can be arbitrarily large.

For example, just take $\sin(nx)$, restricted to $[0,1]$, as $n$ varies over all integers.

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