The graph of a linear operator is closed but not continuous (closed graph theorem)

Question

$\textbf{The question is}$

Let $X:=C^{1}[0,1] $, $ Y:=C^{0}[0,1]$ both equipped with the $\left \| f \right \|:=\begin{matrix} Sup & \left | f(x) \right |\\ x\in [0,1]& \end{matrix}$ and let $\begin{matrix} T:& X&\rightarrow &Y \\ &f &\mapsto &{f}' \\ \end{matrix}$.

Then $G(T):=\left \{ (f,T(f))\mid f \in X \right \}$ is closed and $T$ is not continuous

In this problem we have to $ \left \| f \right \|_X+\left \| g \right \|_Y=:\left \| (f,g) \right \|_{X\times Y}$

$\textbf{This is what I have demonstrated}$

1. if $f(x):=-1\in X$ then $\forall c>0: -c=cf<0= \begin{matrix} Sup & \left | 0 \right |\\ x\in[0,1] & \end{matrix} = \left \| 0 \right \| :=\left \| T(-1) \right \|$

In conclusion $T$ is not continuous

2. Although I don't know if it can be of any use to us, I have already demonstrated that $Y$ is Banach Space, I won't write the details so as not to extend so much.

$\textbf{Attempt to see that G(T) is closed}$

let $\left \{ (f_n,f^{'}_{n}) \right \}$ a sequence in $G(T)$ and be $(f,g)\in X\times Y$ such that $\begin{matrix} (f_n ,f^{'}_{n})&\rightarrow &(f,g) \\ n&\rightarrow &\infty \\ \end{matrix}$

note that $\forall \epsilon >0: \exists N >0 :\forall n\geq N:$

$\left \| f_{n}-f \right \|,\left \| f^{'}_{n}-g \right \|\leqslant \left \| f_{n}-f \right \|+\left \| f^{'}_{n}-g \right \|=\left \| (f_n-f,f^{'}_{n}-g) \right \|_{X\times Y}=\left \| (f_{n}-f^{'}_{n})-(f,g) ) \right \|_{X\times Y}<\varepsilon $

then $ \begin{matrix} f_n& \rightarrow & f\\ n&\rightarrow & \infty \end{matrix}$ and $\begin{matrix} f^{'}_n& \rightarrow & g\\ n&\rightarrow & \infty \end{matrix}$

my idea was to try to get them to $\forall \epsilon >0: \left \| T(f)-g \right \|<\epsilon $ but I think this would demonstrate the continuity of T and that we have already seen that it can not be

$\textbf{My question is}$

How could I prove that $G(T)$ is closed, I would appreciate if you could guide me.

Answer 1

$T(-1)=0$ does not show that $T$ is dis-continuous. If $f_n(x)=\frac {x^{n}} n$ then $f_n \to 0$ uniformly but $Tf_n$ does not converge uniformly. This shows that $T$ is not continuous.

Given $f_n \to f$ uniformly ($f_n, f \in C^{1}([0,1])$) and $f_n \to g$ uniformly we have to show that $g=f'$. For this we have $f_n(x)=f_n(0)+\int_0^{x} f_n'(t)dt$. Taking limit we get $f(x)=f(0)+\int_0^{x} g(t)dt$. Since $g$ is continuous it follows that $f$ is differentiable and $g=f'$.