Is $T$ continuous and has closed graph?

Question

$T:C^1[0,1]\to C[0,1]$ is defined by $T(x(t))=x'(t)+\int_{0}^{t} x(t)\, dt$

Is $T$ continuous and has closed graph?

$x(t)=t^n$ then $T$ is not bounded and hence not continuous and has not closed graph. am I right?

Answer 1

You have $$ \|Tx\|_{C[0,1]} \leq \|x'\|_{C[0,1]} + \|x\|_{C[0,1]} \leq c \|x\|_{C^1[0,1]}, $$ for some constant $c$ that depends on the exact definition of the $C^1$ norm you are using, and so $T$ is continuous and hence has a closed graph.

Here you made a mistake because I think you did not notice that the domain of $T$ is $C^1$, so you can bound $Tx$ in terms of the $C^1$ norm of $x$. In particular, having a derivative of $x$ is not so dangerous.

Another thing to note is that in general an operator can be not continuous but still have closed graph.