Let $f \in L^2([0,1])$, and $T : L^2([0,1]) \to L^2([0,1]),\ f \mapsto Tf$.
For all $x \in [0,1]$, $\displaystyle Tf(x)= \int_{0}^{x}f(u) \mathrm{d}u$.
The goal is to compute $\Vert T \Vert$.
Applying the Cauchy-Schwarz inequality gives : for all $f \in L^2([0,1])$, $\Vert T f \Vert_2 \le \Vert f \Vert_2$. Hence $\Vert T \Vert \le 1$.
Now, I struggle to build a function $f\in L^2([0,1])$ such that : $\Vert T f \Vert_2 = \Vert f \Vert_2$. I tried several constant functions for $f$ but there is always a constant problem. Maybe the bound $1$ is not the right one.
Note that $\displaystyle \Vert Tf \Vert_2=\left(\int_{0}^{1}\left\vert \int_{0}^{x} f(u) \mathrm{d}u\right\vert ^2 \mathrm{d}x \right)^{\frac{1}{2}}$.
Also, if I have determined the adjoint operator $T^*$ (which is not self-adjoint), does the norm of $\Vert T^* \Vert$ can be computed using $\Vert T \Vert$ ?
Thanks in advance !
