I am asked to compute the adjoint of the operator $A: L^2[0,1] \longrightarrow L^2[0,1] $ defined by $$ Af(t)=\int_{t}^{1} t (s-1/2) f(s) ds $$ and then to compute its norm, that is $\lVert A^* \rVert$.
My attempt:
For the first part, we have that for all $f,g \in L^2[0,1]$:
\begin{align*} \langle Af,g \rangle&= \int_{0}^{1} Af(t)g(t) dt= \int_{0}^{1} \left(\int_t^1 t (s-1/2) f(s) ds \right) g(t) dt\\ &= \int_{0}^{1} \int_{0}^{s} t (s-1/2) f(s) g(t) dt\, ds= \int_{0}^{1} f(s) \left( \int_{0}^{s} t(s-1/2)g(t)dt \right) ds \end{align*} and therefore $$ A^{*}g(s)= \int_{0}^{s} t(s-1/2)g(t)dt $$
Now, since $A$ is bounded (by $\lVert k \rVert_{2}$, where $k(t,s)=1_{[t,1]}(s)t(s-1/2) \in L^2([0,1]\times[0,1])$ for example) I know that $\lVert A^* \rVert=\lVert A \rVert$ which led me to trying to compute $\lVert A \rVert$. One natural candidate for $\lVert A \rVert$ is precisely $\lVert k \rVert_{2}=\frac{\sqrt{7}}{ \sqrt{720} }$. I tried to find $f \in L^2[0,1]$ such that $\lVert f \rVert_2=1$ and $\lVert Af \rVert_2=\lVert k \rVert_2=1$ but it wasn't such a good idea and consequently my approach is probably wrong.
Any help or suggestion?
In advance thank you.