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Consider $ X= C^\infty ([0,1],\mathbb{R} )$ and the operator $$ D: (X,\|\cdot\| ) \to (X,\|\cdot\|).$$ given by $$D(x)=x'.$$ (derivation operator).

Why is unbounded, independently of the choice of norm? I can prove this for cases where I know the norm, but why does this hold in the general case?

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    $\begingroup$ limited = bounded? $\endgroup$ – daw Oct 27 at 20:09
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Because the exponential functions $e^{cx}$ with $c\in\mathbb{R}$ exist in $C^\infty([0,1],\mathbb{R})$: $$\|D(e^{cx})\|=\|ce^{cx}\|=|c|\cdot\|e^{cx}\|,$$ and there is no constant $M\geqslant0$ such that this norm is $\leqslant M\cdot\|e^{cx}\|$ for all $c \in \mathbb{R}$.

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