Given an operator $T: L^2[0,1] \to L^2[0,1]$ which is a composition $T = A(id - B)$ where $A$ takes $f(x) \in L^2[0,1]$ to $xf(x)$ and $B$ takes $f(x)$ to $\int^{1}_{0} f(t)dt$.
It is easy for me to see that $T$ is bounded because $\|T\| \leq \|A\| + \|B\| \leq 2$ but how do I calculate the norm of $T$?
Notice that $\|A\|=1$ (see Show that the norm of the multiplication operator $M_f$ on $L^2[0,1]$ is $\|f\|_\infty$). Moreover $$\|(I-B)f\|_2^2=\int_0^1\left(f(x)-\int_0^1f(t)dt\right)^2dx=\|f\|_2^2-\left(\int_0^1f(t)dt\right)^2\leq \|f\|_2^2$$ and the equality holds for $f(x)=2\sqrt{2}(x-1/2)$ with $\|f\|_2=1$. Hence $\|I-B\|=1$.
Now the operator norm is compatible with the composition and $$\|T\|\leq \|A\|\cdot\|I-B\|=1.$$ Finally, $\|T\|=1$ follows by considering the functions $f_n(x)=\sqrt{2n+1}\,x^n$. Then $\|f_n\|_2=1$, and, as $n$ goes to infinity, $$\|Tf_n\|_2^2=\int_0^1x^2\left(f(x)-\int_0^1f(t)dt\right)^2 dx =\frac{2n+1}{2n+3}-\frac{2(2n+1)}{(n+3)(n+1)}+\frac{2n+1}{3(n+1)^2}\to 1. $$
P.S. We may also use $f_n(x)=\sqrt{n}\,1_{[1-1/n,1]}(x)$.