On the space of $L^2([0,1])$ consider the following operator; $\Lambda u(x)= \int_0^xu(s)ds$ , I want to find the adjoint of this operator.
\begin{align}
(\Lambda u , g)
& = \int_0^1\Big(\int_0^xu(s)ds\Big)g(x)dx \\
& = \Big(\int_0^1u(x)dx\Big)\Big(\int_0^1g(x)dx\Big) – \int_0^1 u(x)\Big(\int_0^xg(s)ds\Big)dx
\end{align}
where I did integration by part (is it done correctly?). But I stuck how to proceed for getting $( u , \Lambda^\ast g)$
Best Answer
You almost got it. Since $(\Lambda g)'=g$ and $(\Lambda u)'=u$, we have $$\begin{align}\langle \Lambda u,g\rangle &=\int_0^1\,(\Lambda u)(x)\,g(x)\,\text{d}x \\&=\big((\Lambda u)(x)\,(\Lambda g)(x)\big)\Big|_{x=0}^{x=1}-\int_0^1\,u(x)\,(\Lambda g)(x)\,\text{d}x \\&=(\Lambda u)(1)\,(\Lambda g)(1)-\int_0^1\,u(x)\,(\Lambda g)(x)\,\text{d}x \\&=\left(\int_0^1\,u(x)\,\text{d}x\right)\,(\Lambda g)(1)-\int_0^1\,u(x)\,(\Lambda g)(x)\,\text{d}x \\&=\int_0^1\,u(x)\,(\Lambda g)(1)\,\text{d}x-\int_0^1\,u(x)\,(\Lambda g)(x)\,\text{d}x \\&=\int_0^1\,u(x)\,\Big((\Lambda g)(1)-(\Lambda g)(x)\Big)\,\text{d}x=\langle u,\Lambda^\dagger g\rangle\,, \end{align}$$ where $$(\Lambda^\dagger g)(x):=(\Lambda g)(1)-(\Lambda g)(x)=\int_x^1\,g(s)\,\text{d}s\,.$$