Find the norm of these operators that map $:L_2[0,1] \to :L_2[0,1]$ The first one: 
$$a) A(x(t)) = \int_{0}^{t}x(s)ds$$ 
Here I proved that $\|A(x(t))\|_{L_2[0,1]} \leq \|x(t)\|_{L_2[0,1]}$. Unfortunately I cannot find a function that has the norm of one and the map having the norm of one too.
$$b) \lambda \in (0,1), A(x(t)) = \begin{cases}
x(t), t \leq \lambda \\
0, t > \lambda
\end{cases} t\in [0,1]$$ 
$\|A(x(t))\|_{L_2[0,1]} \leq \|x(t)\|_{L_2[0,1]}$ I got this again, but I am sure the norm is less than one. Closer to $\lambda$. But I dont know how to prove this.
 A: First of all, the proof of the "other direction" is important for motivating an example.  Here's my version:
$$
\|Ax\| = \int_0^1 \left|\int _0^tx(s)\,ds\right|^2 dt \leq 
\int_0^1 \int _0^t |x(s)|^2\,ds\, dt \\
= \int_0^1 \int_s^1 |x(s)|^2 \,dt\,ds = \int_0^1\, (1-s)|x(s)|^2\,ds\\
\leq \int_0^1 |x(s)|^2\,ds
$$
Now, from that second to last step, perhaps you can intuit that for any $x(s)$, we will have $\|Ax\| < \|x\|$.  However, we're ultimately interested in a supremum.  Ultimately, what we really want is to find an $x_\epsilon(t)$ such that $\|A x_\epsilon\|/\|x_\epsilon\| \to 1$ as $\epsilon \to 0^+$. Intuitively, such a function should have most of its "weight" on the left side.
In particular, consider
$$
x_\epsilon(t) = 
\begin{cases}
1/\sqrt{\epsilon} & t < \epsilon\\
0 & t \geq \epsilon
\end{cases}
$$
Calculate $\|x_\epsilon\|^2 = 1$, whereas 
$$
Ax_\epsilon(t) = \begin{cases}
t/\sqrt{\epsilon} & t < \epsilon\\
1/\sqrt{\epsilon} & t \geq \epsilon
\end{cases}
\\
\|Ax_\epsilon\|^2 = (1 - \epsilon)/\epsilon + \int_0^\epsilon (t/\epsilon)^2\,dt
$$
For $b$: in fact, the norm of the operator is $1$. Consider
$$
x(t) = \begin{cases}
1 & t \leq \lambda\\
0 & t \geq \lambda
\end{cases}
$$
Verify that $Ax = x$, so that $\|Ax\| = \|x\|$.  As a linear operator, we might regard this $A$ as an orthogonal projection onto a subspace.
