# Consider the following linear operation

$T: (C^1([0,1]),||.||_u) \rightarrow (C([0,1]),||.||_u)$ , $T(f)=f'$

Let $Gr(T)=\{(x,Tx);x\in X\}$.

To show that $T$ is not bounded consider $f_n(x)= \frac{Sin(n x)}{n}$.

Thus $T(f_n)$ is not converges so is not bounded.but whether $Gr(T)$ is closed?

• math.stackexchange.com/questions/260924/… – Max Oct 20 '15 at 11:18
• Thanks....but I can not to show that Gr(T) is closed.... – m.am Oct 20 '15 at 11:22
• Not convergent does not imply not bounded in general – Vishesh Oct 20 '15 at 11:25
• yes...but for this question is true... – m.am Oct 20 '15 at 11:26
• might the fundamental theorem of calculus help here? This would mean to consider the inverse operator in a way. (btw: sorry for my first comment, i didnt read the answers there good enuff.) – Max Oct 20 '15 at 11:31