# Continuity of the linear transformation $T(x(t))=x'(t)+\int_0^t x(t)\,dt.$

Let $X=C^{1}[0,1]$ with $\|x\|=sup_{[0,1]}(x)+sup_{[0,1]}(x^{'})$ and $Y=C[0,1]$ with supremum norm. Let $$T(x(t))=x'(t)+\int_0^t x(u)\,du.$$ Then $T : X\rightarrow Y$ is

$A.$ Continuous but not closed.

$B.$ Not continuous but closed.

$C.$ Continuous and closed both.

$D.$ Not continuous and not closed.

$\|T(x)\|=\|x'+\int_0^t x(u)\,du\|= \sup\{ |x'+\int_0^t x(u)\,du|\}\leq \sup \{|x'|\} + \sup\{|\int_0^t x(u)\,du|\}\leq sup(x^{'})+sup(x(t))=\|x\|?$ Hence $T$ is bounded and so closed as continuous linear map from Banach space to Banach. Please suggest me. Thanks a lot.

• yes i edited ....thanks..... – neelkanth Aug 16 '16 at 12:07
• This is not correct. The estimate $\sup \{|x'|\} \le \|x\|$ does not hold. – Hans Engler Aug 16 '16 at 12:40
• Are you considering $T : X \rightarrow Y$? – DisintegratingByParts Aug 16 '16 at 14:15
• The problem really specified the suup norm on $C^1$? That's not the usual norm... – David C. Ullrich Aug 16 '16 at 14:18
• @HansEngler I am using $sup(f+g)\leq sup(f)+sup(g).$ – neelkanth Aug 16 '16 at 15:22

The norms on your spaces $X=C^1[0,1]$ and $Y=C[0,1]$ are \begin{align} \|x\|_{X} & = \sup_{t\in[0,1]}|x'(t)|+\sup_{t\in[0,1]}|x(t)| \\ \|y\|_{Y} & = \sup_{t\in[0,1]}|x(t)|. \end{align} Both $X$ and $Y$ are complete spaces.
The operator $T : X\rightarrow Y$ is bounded because \begin{align} \|Tx\|_{Y} & \le \sup_{t\in[0,1]}|x'(t)|+\sup_{t\in[0,1]}\left|\int_{0}^{t}x(u)du\right| \\ & \le \sup_{t\in[0,1]}|x'(t)| + \int_{0}^{1}|x(u)|du \\ & \le \sup_{t\in[0,1]}|x'(t)| + \sup_{u\in[0,1]}|x(u)| = \|x\|_{X}. \end{align} Therefore, $T : X\rightarrow Y$ is continuous. $T$ is also closed because it is continuous.