Show that $\sup_{\|f\| \le 1}\|T(f)\|=\dfrac{1}{\sqrt{2}}$, where $(T(f))(t) = \int_0^t f(s)ds$ and $\|f\|=(\int_0^1 |f(t)|^2dt)^{1/2}$. $f \in L^2[0,1]$, the vector space of all complex valued Lebesgue measurable functions that are square integrable on $[0,1]$, with norm:
$\|f\|=(\int_0^1 |f(t)|^2dt)^{1/2}$.
$T:L^2[0,1] \rightarrow L^2[0,1]$ is a linear operator:
$(T(f))(t) = \int_0^t f(s)ds$ for every $t\in [0,1]$
Show that $\sup_{\|f\| \le 1}\|T(f)\|=\dfrac{1}{\sqrt{2}}$.

This is what I've tried:
$\sup_{\|f\| \le 1}\|T(f)\|^2=\sup_{\|f\| \le 1} \int_0^1|(T(f))(t)|^2dt = \sup_{\|f\| \le 1} \int_0^1|\int_0^t f(s)ds|^2dt 
=\int_0^1 \sup_{\|f\| \le 1} |\int_0^t f(s)ds|^2dt$
then I should make this equal to $\int_0^1 tdt$ by showing $\sup_{\|f\| \le 1} |\int_0^t f(s)ds|^2=t$ and got stuck
 A: Rather than working with the $\sup$ throughout, it will be sufficient to show that
1) For all $f$, $||T(f)||^2 \leq \dfrac{1}{2} ||f||^2$
2) $\exists f : ||T(f)|| = \dfrac{1}{\sqrt 2} ||f||$
So, in working with 1), write out the definitions and conclude
$$
||T(f)||^2 = \int_0^1 \left|\int_0^t f(s) \,ds\right|^2\,dt = \int_0^1 \left|\int_0^1 \chi_{[0,t)}(s)f(s) \,ds\right|^2\,dt \quad\quad (*)
$$
where 
$$
\chi_{[0,t)}(s) = \begin{cases} 1 &: 0 \leq s < t \\
0 &: \textrm{otherwise} \end{cases}
$$
Now, use Cauchy-Schwarz
$$
\left|\int_0^1 \chi_{[0,t)}(s)f(s) \,ds\right| \leq \int_0^1 \left|\chi_{[0,t)}(s)f(s)\right| \,ds \leq ||\chi_{[0,t)}||\cdot||f||.
$$
Next, calculate $||\chi_{[0,t)}||$, then put all the pieces back together in $(*)$, and you are pretty much there.
I'll leave part 2) up to you.
Additionally, it's worth mentioning that this 'trick' of turning limits of integration into a characteristic function (e.g. $\int_0^t \to \int_0^1 \chi$) is a common and useful approach, and you should add it to your collection of problem-solving techniques in analysis.
