Prove that $S:F \to \mathcal{L}(E,F)$ is a topological isomorphism

Question

Let $E$ and $F$ be normed spaces, $E \neq \{ 0 \}$ and $x_0 \in E \backslash \{ 0 \}$, $x_0 \in E'$ such that $x_0'(x_0)=1$. Prove that the function $S: F \to \mathcal{L}(E,F)$, $S(y)=T_y$ defined as $T_y(x)=x_o'(x)y$, is a topological isomorphism and that $S(F)$ is closed.

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I've tried to prove it finding an inverse and then proving that it is continuous because most of the theorems I know are specific to the case where $E$, $F$ or both are Banach spaces, but I wasn't able to get to anything.

Any help would be greatly appreciated.

Answer 1

$$\begin{align}\|S(y_1)-S(y_2)\|_{\mathcal{L}(E,F)}&=\|x_0'(\cdot)y_1-x_0'(\cdot)y_2\|_{\mathcal{L}(E,F)}\\&=\sup_{0\neq x\in E}\frac{\|x_0'(x)y_1-x_0'(x)y_2\|_{F}}{\|x\|_{E}}\\&=\sup_{0\neq x\in E}\frac{|x_0'(x)|}{\|x\|_{E}}\cdot \|y_1-y_2\|_{F}\\&=\|x_0'(\cdot)\|_{E'}\cdot\|y_1-y_2\|_{F}\end{align}$$

Therefore

$$\|S(y_1)-S(y_2)\|_{\mathcal{L}(E,F)}=\|x_0'(\cdot)\|_{E'}\cdot\|y_1-y_2\|_{F}$$

This shows that $S$ is continuous. Linearity and injectivity you probably can do.

The very same equation shows that $S^{-1}:S(F)\to F$ is continuous. just replace $y_1$ and $y_2$ by $S^{-1}(T_{y_1})$ and $S^{-1}(T_{y_2})$ respectively and pass the $\|x_0'(\cdot)\|_{E'}$ dividing to the other side.

$$\|x_0'(\cdot)\|_{E'}^{-1}\cdot\|T_{y_1}-T_{y_2}\|_{\mathcal{L}(E,F)}=\|S^{-1}(T_{y_1})-S^{-1}(T_{y_2})\|_{F}$$