I come across with the following question: Let $X$ and $Y$ be normed spaces over the same field and $T:X \rightarrow Y$ be a linear isometry, with $\|Tx\|=\|x\| \, \forall x \in X$, then the adjoint operator, $T^* : Y^* \rightarrow X^*$ defined by $T^*(f) = f(Tx) \, \forall x \in X$ is surjective.
I would like to proceed as following: As $T$ is bijective between $X$ and $T(X)$, we can define for each $g \in X^{*}$, $f = g \circ T^{-1} \in T(X)^*$ and extend such definition over $Y^*$.
Can I get a small hint from that? Many thanks
You are right. Consider $g \circ T^{-1} \in T(X)^*$, which is bounded, as $T$ is an isometry. Hence, by Hahn-Banach, it can be extended to and $f \in Y^*$. We have $$ (T^*f)x = f(Tx) = (g \circ T^{-1})(Tx) = g(x) $$ for $x \in X$. Hence $T^*f = g$.