Exercise 2.6.17

If $W$ is $T$-invariant, we have that $T(W)\subset W$. Let $f$ be a functional in $W^0$. We can check $T^t(f)=\mathit{fT}$ is an element in $W^0$ since $T(w)\in W$ by the fact that $T$-invariant and thus $f(T(w))=0$.

For the converse, if $W^0$ is $T^t$-invariant, we know $T^t(W^0)\subset W^0$. Fix one $w$ in $W$, if $T(w)$ is not an element in $W$, by Exercise 2.6.13(b) there exist a functional $f\in W^0$ such that $f(T(w))\ne 0$. But this means $T^t(f)(w)=\mathit{fT}(w)\ne 0$ and hence $T^t(f)\notin W^0$. This is a contradiction. So we know that $T(w)$ is an element in $W$ for all $w$ in $W$.