Find norm of the operator $A: C[-1,1] \to C[-1,1]$, $(Ax)(t) = \int_{-1}^t x(s) ds - \int_0^1 s x(s) ds$ Consider the following operator $A: C[-1,1] \to C[-1,1]$, $(Ax)(t) = \int_{-1}^t x(s) ds - \int_0^1 s x(s) ds$. What's its norm?
It's easy to show that $\Vert Ax \Vert \le \frac52 \Vert x \Vert \,\, \forall x \in C[-1,1]$. So $\Vert A \Vert \le \frac52$. But it seems that it's too inaccurate. I suspect that $\Vert Ax \Vert \le \frac32 \Vert x \Vert$ which turns into equality for $x(t) = 1$.
 A: If you have an operator  $T:f\longmapsto \int_{-1}^1g(t,s)f(s)\,ds$, with $g$ good enough (say, so that $g_t$ below is continuous), then
$$\tag1
\|T\|=\sup\{g_t:\ t\in[-1,1]\},
$$
where 
$$
g_t=\int_{-1}^1|g(t,s)|\,ds.
$$
Indeed, we have 
$$
\|Tf\|=\sup_t\left|\int_{-1}^1g(t,s)\,f(s)\,ds \right|\leq\|f\|\,\sup_t\int_{-1}^1|g(t,x)|\,ds.
$$
Given $\varepsilon>0$, there exists $t_0$ such that $g_{t_0}+\varepsilon>\sup\{g_t:\ t\}$. Now choose $f\in C[-1,1]$ such that $\|f\|=1$ and $Tf(t_0)+\varepsilon>g_{t_0}$. Then 
$$
\|T\|+2\varepsilon\geq\|Tf\|+2\varepsilon\geq Tf(t_0)+2\varepsilon>g_{t_0}+\varepsilon>\sup\{g_t:\ t\}. 
$$
As $\varepsilon$ was arbitrary, we have the reverse inequality, and thus equality. 
In the case at hand, we have 
$$
g(t,s)=1_{[-1,t]}(s)-s\,1_{[0,1]}(s).
$$
Then 
$$
\int_{-1}^1|g(t,s)|\,ds=\int_{-1}^1|1_{[-1,t]}(s)-s\,1_{[0,1]}(s)|\,ds.
$$
When $t\leq0$, this is 
$$
\int_{-1}^1 1_{[-1,t]}(s)+s\,1_{[0,1]}(s)|\,ds
=t+1+\int_0^1 s\,ds=t+\frac32.
$$
When $t>0$, it is 
$$
\int_{-1}^1 1_{[-1,0]}(s)+(1-s)\,1_{[0,t]}+s\,1_{[t,1]}(s)|\,ds
=1+t-\frac{t^2}2+\frac12-\frac{t^2}2=t+\frac32-t^2.
$$
The maximum $t$ will then be $t=1/2$, which gives 
$$
\|T\|=\frac32+\frac14=\frac74.
$$
