Put $L^2=L^2(0,\infty)$ relative to Lebesgue measure, and the Cèsaro operator $C$ is defined as follows: $$(Cf)(s)=\frac{1}{s}\int_0^s f(t)dt$$ we can find its adjoint operator: \begin{align*} \langle Cf,g\rangle=&\int_0^{\infty}\left(\frac{1}{s}\int_0^s f(t)dt\right)\overline{g(s)}ds\\ &=\int_0^{\infty}\left(\int_t^{\infty} \frac{1}{s}f(t)\overline{g(s)}ds\right)dt\\ &=\int_0^{\infty}f(t)\left(\int_t^{\infty} \frac{1}{s}\overline{g(s)}ds\right)dt\\ \end{align*} Hence the adjoint is given by $$(C^*f)(t)=\int_t^{\infty} \frac{1}{s}f(s)ds$$ How can I justify the change of the integration limits in the inner integral in the second equality?
I leave the link from where I obtained the proof: https://desvl.xyz/2021/07/08/Cesaro-operator-in-L2/
We write \begin{align*} \int_0^\infty \frac{1}{s} \left( \int_0^s f(t) dt \right) \overline{g(s)} ds = \int_0^\infty \int_0^\infty \frac{1}{s} 1_{(0,s)}(t) f(t) \overline{g(s)} dt ds. \end{align*} Now we would like to change the order of integration. In order to do so, we use the Fubini-Tonelli theorem. In order to do so, we need to show that the iterated integral with absolute values is finite. Using Cauchy-Schwarz and Hardy's inequality we get \begin{align*} \int_0^\infty \left(\frac{1}{s} \int_0^s \vert f(t) \vert dt \right) \vert g(s) \vert ds &\leq \left(\int_0^\infty \left\vert \frac{1}{s} \int_0^s \vert f(t) \vert dt \right\vert^2 ds \right)^{1/2} \left( \int_0^\infty \vert g(s) \vert^2 ds \right)^{1/2} \\ &\leq 2 \Vert f \Vert_2 \Vert g \Vert_2<\infty. \end{align*} After changing the order of integration we get \begin{align*} \int_0^\infty \frac{1}{s} \left(\int_0^s f(t)dt\right) \overline{g(s)} ds = \int_0^\infty f(t) \left(\int_0^\infty \frac{1}{s} 1_{(0,s)}(t) \overline{g(s)} ds \right). \end{align*} We are left to get rid of the indicator function and adjust the lower limit of integration. For this we note that \begin{align*} \int_0^\infty \frac{1}{s} 1_{(0,s)}(t) \overline{g(s)} ds &= \int_0^t \frac{1}{s} \underbrace{1_{(0,s)}(t)}_{=0} \overline{g(s)} ds + \int_t^\infty \frac{1}{s} \underbrace{1_{(0,s)}(t)}_{=1} \overline{g(s)} ds \\ &= \int_t^\infty \frac{1}{s} \overline{g(s)} ds. \end{align*}