adjoint of an operator. on $L^2(0,1)$, $Bf(x)=\int_0^x f(t)dt$

Question

I see that the above operator is bounded.

I ended up with an argument to calculate the adjoint as follows,

$$ <f,Bg>=\int_0^1\overline{f(x)} \int_0^xg(t)\,dt\,dx $$

I see $f(x)$ as the derivative of $\frac{d}{dt}\int_0^t f(x)\,dx$ and use integration by parts in above equation and get

$$ <f,Bg>=-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\big[\int_0^xg(t)\,dt \,\int_0^x\overline{f(t)}\,dt\big]_0^1 $$

So domain of $B^*$ is $f\in L^2(0,1)$ such that $\int_0^1\overline{f(t)}\,dt=0$

But B is a bounded operator, domain of adjoint should be whole space $L^2(0,1)$.

I do not know is the mistake?

Please help

Answer 1

After integrating by parts, you get $$ \langle f,Bg\rangle =-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\big[\int_0^xg(t)\,dt \,\int_0^x\overline{f(t)}\,dt\big]_0^1 $$ Continuing this calculation (inserting the limits, noting that $x=0$ gives nothing, and renaming the integrating variable in the $g$ integral), we get $$ \begin{aligned} \langle f,Bg\rangle &=-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\int_0^1g(x)\,dx \,\int_0^1\overline{f(t)}\,dt\\ &=\int_0^1g(x)\biggl[\int_0^1\overline{f(t)}\,dt-\int_0^x \overline{f(t)}\,dt\biggr]\,dx\\ &=\int_0^1g(x)\int_x^1\overline{f(t)}\,dt\,dx\\ &=\langle Sf,g\rangle, \end{aligned} $$ where $$ (Sf)(x)=\int_x^1 f(t)\,dt. $$