Every map $G$ from $Y^* \to X^*$ can be expresed as $T^*$ where $T: X \to Y$ [duplicate]

Solution 1:

If $S=T^*$, and $f_j\to f$ weak$^*$ in $Y^*$, then for any $x\in X$ $$ Sf_j(x)=f_j(Tx)\to f(Tx)=T^*f(x)=Sf(x). $$ So $S$ is weak$^*$-continuous.

Conversely, assume that $S$ is weak$^*$-continuous. We know what our $T$ should satisfy if it exists: $Sf(x)=f(Tx)$. So let us use this to define $T$.

For any $x\in X$, consider the functional on $Y^*$ given by $\alpha_x:f\longmapsto Sf(x)$. By hypothesis this is weak$^*$-continuous; the weak$^*$-continuous functionals are bounded, so $\alpha_x\in Y^{**}$. But, also, a weak$^*$-continuous functional on the dual is in the predual. So there exists $y\in Y$ with $\alpha_x(f)=f(y)$ for all $f\in Y^*$. Define $Tx=y$. Note that $y$ is unique because $Y^*$ separates points in $Y$. From there we deduce that $T$ is linear. Finally, we have that $S $ is bounded; indeed, if $f_j\to f $ in $Y^* $ and $Sf_j\to g $ in $X^*$ (both in norm), then $f_j\to f $ in the weak$^*$ topology, so $Sf_j\to Sf $ weak$^*$; thus $Sf=g $ and $S $ is bounded by the Closed Graph Theorem. Then \begin{align} \|Tx\|&=\sup\{|f(Tx)|:\ f\in Y^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|Sf(x)|:\ f\in Y^*,\ \|f\|=1\}\leq\|S\|\,\|x\|. \end{align}So $T$ is bounded with $\|T\|\leq\|S\|$.