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Let $X = C ^ \infty ([0,1] , \mathbb R )$ and let $\|\cdot\|$ be any norm of $X$. Define the operator $T:X\to X$ by $T(f) = \frac{df}{dx}$.

Show that $T$ is not a continuous linear operator from $( X , \|\cdot\| )$ into $(X , \|\cdot\| )$,

Hint : Use the function $f_a (x) = \exp(ax) , a \in \mathbb R$.

My answer is :

$T$ is discountinuous if there is $f_a (x) , a \in \mathbb R$, $X = C ^ \infty ([0,1] , \mathbb R )$ such that $\|Tf_a\| \to \infty$ as $a \to \infty$.

$\|Tf_a\|= \|T(e^{ax}) \| = \|a e^{ax} \| \to \infty$ as $a \to \infty$.

Thus $T$ is not continuous.

Is my answer true?

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Assume that $T$ is continuous. Then $||Tf|| \le ||T|| \cdot ||f||$ for all $f \in X$.

Since $Tf_a=af_a$ we get $|a| ||f_a||=||Tf_a|| \le||T|| \cdot ||f_a||$ , hence

$$||T|| \ge a.$$

This holds for all $a \in \mathbb R$, which is absurd.

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There should be the constraint that $\left\lVert f_a\right\rVert$ should be bounded or equal to one for example (otherwise we could take $f_a=a\cdot g$ where $g\neq 0$). But we can choose $g_a:=f_a/\left\lVert f_a\right\rVert$. The norm of $g_a$ is one, that of $Tg_a=g'_a$ is $\left\lvert a\right\rvert$.

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    $\begingroup$ A small misprint: the norm of $Tg_a$ is, generally, $\lvert a \rvert$, not $a$ (the OP assumes $a \in \mathbb{R}$). $\endgroup$ – user539887 Apr 17 '18 at 9:20
  • $\begingroup$ @user539887 Right. $\endgroup$ – Davide Giraudo Apr 17 '18 at 9:31

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