Define $T:X\rightarrow L^{p}[0,1]$ where $Tf=f'$. Here the function $f$ is in $C^{\infty}[0,1]$ and $f(0)=0$. How do I show that this $T$ is an injective linear operator that is not bounded? Correct me if I am wrong, the prime here refers to the differential operator.

To show that it is linear is pretty straightforward. But how is it that $T$ is injective? One to one?

Also, how do I show that it is unbounded? Do I come up with a specific function $f$?

Lastly, what about the inverse of $T$? Is it unbounded too?

Just a note: I have read some articles and it seems like letting $f_{k}=sinkx$ will do the job but the norm is infinity norm. Here $p$ is finite.

  • $\begingroup$ Can you specify what $X$ and the norm on $X$ is? Also $T$ is not injectiv, since for every constant $c\in\mathbb{R}$ you have $(f+c)'=f'$. $\endgroup$ – humanStampedist Mar 1 '18 at 9:22
  • $\begingroup$ $1\leq p<\infty$. $X$ is a subspace of $L^{p}[0,1]$ $\endgroup$ – LanaDR Mar 1 '18 at 9:24
  • $\begingroup$ How can $X$ be $L^p$ if we are computing a derivate? From what you wrote it looks like $X=C^\infty$ equipped with the uniform norm. $\endgroup$ – Tommaso Seneci Mar 1 '18 at 9:25
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    $\begingroup$ @humanStampedist It is injective because of the condition $f(0)=0$. For the unboundedness a hint would be to look at $f(x)=x^n$. $\endgroup$ – Maik Pickl Mar 1 '18 at 9:25
  • $\begingroup$ @Maik Picki: Thanks, totally overlooked that. $\endgroup$ – humanStampedist Mar 1 '18 at 9:27

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