TITLE: Prove $T: X =C^1[0,1] \rightarrow \mathbb R: Tf=\int_{0}^{1} f(t) dt$ is continuous
QUESTION [0 upvotes]: Let $X(C^1[0,1],\lvert\lvert.\rvert\rvert)$  with the norm $\lvert\lvert f\rvert\rvert=\lvert 
f(0)\rvert+\sup_{t\in[0,1]}\lvert f'(t)\rvert$ 

To prove the continuity of the operator I proved it's bounded.
$|Tf|=|\int_{0}^{1} f(t) dt|\le \int_{0}^{1} |f(t)| dt\le \sup_{t \in [0,1]}|f(t)|$
so is bounded and then continuous?
May I ask how to calculate $\lvert\lvert T\rvert\rvert$?
Thank you in advance!

REPLY [1 votes]: From the context, I assume $Tf=\int_0^1f(t)dt$.
A linear functional is continuous $\iff$ $\sup_{||f||=1}|Tf|\leq C$.
Furthermore, we define $||T||=\sup_{||f||=1}|Tf|$.
For the continuity, we you have already done some work. Just note that 
$$
|f(t)|=|\int_0^t f'(s)ds+f(0)| \leq |f(0)|+\sup_{t \in [0,1]}|f'(t)|
$$
and therefore, $\sup_{t \in [0,1]}|f(t)| \leq ||f||$.
Combining with your estimate, we have:
$$
\sup_{||f||=1}|Tf|\leq ||f||\leq 1
$$ 
This estimate is sharp. Indeed, consider the function $f(x)=1$. Then 
$$
T(f)=1=||f||
$$
and as a result, this functions attains the surpremum.
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