# Under what norm $F: X \to X'$ is continuous?

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

• 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. – David C. Ullrich Jul 11 '19 at 1:00
• @DavidC.Ullrich That's why I consider a room for considering a subspace of AC, Like$Y= C^{\infty} [0,1]$ . – Red shoes Jul 11 '19 at 3:48
• @DavidC.Ullrich I will Edit the question, I only care the domain of $F$ be $X$. – 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.
• Thanks ! By one dimension you mean $n =1$ ? does this also work for for any $n$ if consider the components of $f$? – Red shoes Jul 11 '19 at 21:24
• No, one dimension means $[0,1]\subset \mathbb R^1$. – gerw Jul 12 '19 at 5:07