Consider the sesquilinear form $$B(f,g)=\int_0^1\bigg(\int_0^xf(t)^*dt\bigg)\bigg(\int_0^xg(t)dt\bigg)dx$$ in $L^2(0,1)$. Show that it is bounded.
Let $q(f)=B(f,f)$ be the associated quadratic form. Then $\|B\|\le2\|q\|$, so it suffices to show that $|B(f,f)|\le C\|f\|$ for some constant $C$. There is a hint to use integration by parts so I did so and got: $$B(f,f)=\int_0^1f(x)\int_x^1\int_0^y(f(t))^*dtdydx$$ But I'm not seeing how this helps. Perhaps I should be applying some inequalities first and then integrating by parts? Any help is appreciated.
$|B(f,f)|≤ C\|f\|$ is only possible if $B=0$. Boundedness for a sesqui-linear form means $|B(f,g)|≤ C\|f\|\ \|g\|$. To see that your form satisfies this condition note that : $$|B(f,g)| ≤\int_0^1 \int_0^x |\overline{f(t)}|dt \int_0^x |g(t)| dt \ dx ≤ \int_0^1 |f(t)| dt \int_0^1 |g(t)| dt$$ now $\int_0^1 |f(t)| dt = \langle |f| ,1\rangle$ where $1$ is the constant function. So by Cauchy Schwarz this is smaller than $\|f\|\cdot \|1\|$. In this case you have $\|1\|=1$ Hence: $$|B(f,g)| ≤ \|1\|^2 \ \|f\|\ \|g\|=\|f\|\ \|g\|$$ giving boundedness.