Let $X = AC[0,1]$ be the space of all absolutely continuous functions from $[0, 1]$ to $\mathbb{R}^n$, and let $X'=\text{The space of all Lebesgue integrable functions}$.

Consider the linear function $F: X \to X'$ defined by $F(f)=f'$. In order to $F$ be well-defined we only consider $f'$ where it exists (It almost every where exists on $[0,1]$). Clearly $F$ is linear, my question is that, is there any interesting norm on $X$ or a subspace $Y \subset X$ with a suitable norm such that it turns $F$ a continuous function ?

I'm not looking for trivial norms. This operator frequently appear in optimal control, and the motivation of this question is here

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    $\begingroup$ If $X=AC[0,1]$ then $F(f)=f'$ does not define a map from $X$ to $X$ in the first place. The derivative of an AC function can be any Lebesgue integrable function; it certainly need not be AC. $\endgroup$ – David C. Ullrich Jul 11 '19 at 1:00
  • $\begingroup$ @DavidC.Ullrich That's why I consider a room for considering a subspace of AC, Like$ Y= C^{\infty} [0,1]$ . $\endgroup$ – Red shoes Jul 11 '19 at 3:48
  • $\begingroup$ @DavidC.Ullrich I will Edit the question, I only care the domain of $F$ be $X$. $\endgroup$ – Red shoes Jul 11 '19 at 3:53

In one dimension, you have $AC[0,1] = W^{1,1}(0,1)$, which is a Sobolev space. Moreover, your space $X'$ coincides with $L^1(0,1)$. Thus,

$$\| F(f) \|_{L^1} = \| f' \|_{L^1} \le \| f \|_{W^{1,1}}.$$ Therefore, $F$ is bounded with these natural norms.

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  • $\begingroup$ Thanks ! By one dimension you mean $n =1$ ? does this also work for for any $n$ if consider the components of $f$? $\endgroup$ – Red shoes Jul 11 '19 at 21:24
  • $\begingroup$ No, one dimension means $[0,1]\subset \mathbb R^1$. $\endgroup$ – gerw Jul 12 '19 at 5:07

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