Show that $T$ is closed but not continuous. Let $T: C^1[0, 1] \rightarrow C[0, 1]$ be a linear transformation, both with sup norm, defined by $Tf(s)=f'(s)+\int _0^s f(t)dt$. Show that $T$ is closed but not continuous.
If we take $x_n(t)=t^n$ then $T $ is unbounded but how to show $T$ is closed. Please help
 A: To show that $T$ is a closed operator consider a sequence $(f_n)_{n \in \mathbb{N}} \subset C^1[0,1]$ such that $f_n \to f$ in $(C^1[0,1], \| \cdot \|_\infty)$ and $Tf_n \to g$ in $(C[0,1,], \| \cdot \|_\infty)$. We need to show that $Tf = g$. I am going to assume that you know that this is a equivalent condition for $T$ being closed. If you do not know about it please tell me about your definition of a closed operator.
First of all we can deduce that
$$r_n(s):= \int_0^s f_n(x) \mathrm{d}x \to \int_0^s f(x) \mathrm{d}x $$
unfirfomly in $[0,1] $ due to the uniform convergence of the sequence $(f_n)_{n \in \mathbb{N}}$ and the conintuity of the so called Volterra integral operator. Since $T f_n \to g$ uniformly we conclude that 
$$[T f_n](s) -r_n (s) = f_n'(s) \to g(s)- \int_0^s f(x) \mathrm{d}x$$
uniformly in $[0,1]$. To prove $Tf=g$ it suffices to show that $g(s)-\int_0^s f(x) \mathrm{d}x = f'(s)$ holds for all $s\in [0,1]$. 
Lets recap. We know that $f_n \to f$ uniformly, that $f_n'$ converges uniformly to some function and that every $f_n$ is continously differentiable. Using the Theorem on slide 4 we conclude that $f_n'$ indeed converges to $f'$ which in return implies that the eaulity $f'(s) = g(s)-\int_0^s f(x) \mathrm{d}x $ holds for all $s \in [0,1]$. This shows that $Tf=g$ and thus that $T$ is a closed operator.
