Question:
Let $C^{'}[0,1]$ and $C[0,1]$ be endowed with sup norm. Define $T:C^{'}[0,1]\to C[0,1]$ by $$Tf=f^{'}\text{ for each }f\in C^{'}[0,1]$$ Where, $'$, indicates differentiation.
Prove that $T$ is a linear map with closed graph but it is not bounded.
$Proof:$
For linearity:
Let $f_{1}$, $f_{2}\in C^{'}[0,1]$ $\alpha,\beta$ scalars, then $$T(\alpha f_{1}+\beta f_{2})=(\alpha f_{1}+\beta f_{2})^{'}=\alpha f_{1}^{'}+\beta f_{2}^{'}=\alpha Tf_{1}+\beta Tf_{2}.$$
For the closer of $T$,
Let ${f_{n}}$ be a sequence in $C^{'}[0,1]$ such that $f_{n}\to f$ and $Tf_{n}=f_{n}^{'}\to y$. Then
$\|Tf_{n}-y\|=\sup_{t\in [0,1]}|T(f_{n})(t)-y(t)|=\sup_{t\in [0,1]}|T(f_{n}^{'})(t)-y(t)|\to 0$ as $n\to \infty$
Thus, the convergence is uniform and $y(t)=\lim_{n\to \infty}f_{n}^{'}(t)$. Since the convergence is uniform $f^{'}(t)=y(t)$ for all $t\in [0,1]$. Thus, $f\in C^{'}[0,1]$ and $Tf=y$ so $T$ is closed.
To show $T$ is not bounded, take $f_{n}(t)=t^{n}$, then $\|f_{n}\|=\sup_{t\in [0,1]}|t^{n}|=1$ and $f^{'}_{n}(t)=nt^{n-1}$ so that $\|Tf_{n}\|=\sup_{t\in [0,1]}|nt^{n-1}|=n$.
Thus, $T$ is not bounded. This implies $T$ is not continuous.
My question here is, does this example contradict the Closed Graph theorem? Which says: If you have two Banach spaces $X$ and $Y$ and $T$ a linear map from $X$ to $Y$ such that the graph of $T$ is closed. Then $T$ is continuous.