compute the adjoint operator

Question

Suppose there is a bounded linear operator $T:L^2[-1,1]\to L^2[-1,1]$ given by $Tf(t)=\int_{-1}^0f(s)ds+(\int_{-1}^1f(s)ds)t^2$. We need to compute the adjoint of the operator $T$. When I calculate the integral, I computed as this. \begin{aligned}\langle Tf(t)g(t)\rangle&=\int_{-1}^1[\int_{-1}^0f(s)ds+(\int_{-1}^1f(s)ds)t^2]g(t)dt\\&=\int_{-1}^1\int_{-1}^0f(s)dsg(t)dt+\int_{-1}^1\int_{-1}^1f(s)dst^2g(t)dt\\&=\int_{-1}^0\int_{-1}^1f(s)g(t)dtds+\int_{-1}^1\int_{-1}^1f(s)t^2g(t)dtds.\end{aligned} I don't know whether my calculation is correct and I don't know how to compute the next step.

Answer 1

I think it helps to combine the terms differently

\begin{align*} Tf(t) &= (1+t^2)\int_{-1}^0 f(s) ds + t^2\int_{0}^1 f(s) ds \\ \end{align*}

We then have

\begin{align*} <Tf,g> &= \int_{-1}^1 \left( (1+t^2)\int_{-1}^0 f(s) ds + t^2\int_{0}^1 f(s) ds \right) g(t) dt \\ &= \int_{-1}^0 f(s) \int_{-1}^1 (1+t^2)g(t) dtds + \int_{0}^1 f(s) \int_{-1}^1 t^2g(t) dtds \end{align*}

and we can define

\begin{equation*} h(s) = \begin{cases} \int_{-1}^1 (1+t^2)g(t) dt,& s<0\\ \int_{-1}^1 t^2g(t) dt, & s\geq 0 \end{cases} \end{equation*}

Adding Clarification

To finish, we see that

\begin{align*} <Tf,g> &= \int_{-1}^0 f(s) h(s) ds + \int_{0}^1 f(s) h(s) ds \\ &= \int_{-1}^{1}f(s)h(s)ds \end{align*}

so

\begin{equation*} T^*g(s) = \begin{cases} \int_{-1}^1 (1+t^2)g(t) dt,& s<0\\ \int_{-1}^1 t^2g(t) dt, & s\geq 0 \end{cases} \end{equation*}