Let $X := C^1 [0,1]$ and define $||f|| := |f(0)| + \sup_{0\le t \le 1} |f'(t)|$. Now, consider the operator $T: X \to \mathbb{R}$ defined as $$Tf = \int_0^1 f(t)dt$$ Show that T is continuous.
I know this should be proved with the closed graph theorem, but I want to ask if this alternative proof holds too:
Let's show $T$ is bounded:
$$ \|T\| = \sup_{\|f\| = 1} |Tf| = \sup_{\|f\| = 1} \left| \int_0^1 f(t)\,dt \right|$$
Now, for all $f \in C^1 [0,1]$, we have $\|f\| = 1 \iff1 - |f(0)| = \sup_{0\le t \le 1} |f'(t)|$
We can deduce that $$0 \le \sup_{0\le t \le 1} |f'(t)| \le 1$$
So every function $f$ such that $\|f\| =1$ can't tilt more than the function $f(x) = x$.
$$ \Rightarrow \| T \| = \sup_{\|f\| = 1} \left| \int_0^1 f(t)\,dt \right| \le \left| \int_0^1 t \,dt \right| = \frac{1}{2}$$
Let $f\in X$, and let $M=\sup_{0\leq t\leq 1}|f'(t)|$. Then:
$|T(f)|=|\int_0^1 f(t)dt|=|\int_0^1 (f(t)-f(0))dt+\int_0^1 f(0)dt|=|\int_0^1 (f(t)-f(0))dt+f(0)|\leq$
$\leq |f(0)|+\int_0^1 |f(t)-f(0)|dt\leq |f(0)|+M\int_0^1 tdt=|f(0)|+\frac{M}{2}\leq |f(0)|+M=||f||$
We use the mean value theorem here, for every $t$ we have $|f(t)-f(0)|\leq M|t-0|$.