How can I show that $f\mapsto f'$ is a closed map?

Question

Let $X=C^1([0,1])$ and $Y=C([0,1])$ both equipped with $\|f\|=\max |f|$. Then consider $$D:X\rightarrow Y:~~f\mapsto f'$$ I want to check that $D$ is a closed map.

By definition we need to show that the graph is closed, i.e. $G(D)=\{(f,Df):f\in X\}$ is closed in $X\times Y$. So let $(f_n,Df_n)\rightarrow (f,g)\in X\times Y$. I want to show that $(f,g)\in G(D)$ i.e. that $Df=g$. Now $$\begin{align}\|Df-g\|&=\|Df-Df_n+Df_n-g\|\\&\leq\|Df-Df_n\|+\|Df_n-g\|\\&\leq\|D\|\|f-f_n\|+\|Df_n-g\| \end{align}$$ Now we know $$f_n\rightarrow f \Leftrightarrow \|f_n-f\|\rightarrow 0$$$$Df_n\rightarrow g\Leftrightarrow \|Df_n-g\|\rightarrow 0$$so it remains to check that $||D||<\infty$. $$\|D\|=\sup_{\|f\|\leq 1}\|Df\|=\sup_{\|f\|\leq 1}\|f'\|=\sup_{\|f\|\leq 1}\max|f'|$$ but now since $f'$ is continuous and $[0,1]$ is compact we know that the maximum exists so let me denote $\max|f'|=M$, then $$\|D\|=\sup_{\|f\|\leq 1}M=M<\infty$$ hence as $n\rightarrow \infty$ we get $\|Df-g\|=0$ and therefore $D$ is closed.

Is this correct?

Answer 1

You have to adapt the proof of the following classic theorem

If $(f_n)$ converges pointwise towards $f$ and $f_n'$ converges uniformally towards $g$ then $f$ is $C^1$ and $f' = g$

Let $t \geq 0$

$(f(t)-f(0))-(f_n(t)-f_n(0)) = \int_0^t f'(s)ds - \int_0^t f_n'(s)ds$

And $n \to \infty$ gives :

$0 = \int_0^t f'(s)ds - \int_0^t g(s)ds$ (since $f_n'$ converges uniformally towards $g$)

Which gives the result by differentiating the equality