Let $X$ be a Banach space and suppose $T:X^*\to X^*$ is a linear mapping. If $T$ is norm-norm continuous, i.e. continuous from the normed space $X^*$ into the normed space $X^*$, is it also continuous when $X^*$ is equipped with its weak-star topology (in both the domain and codomain)?
I am a novice with respect to weak topologies and weak-star topologies, so any help is appreciated. Thank you.
No. Let $X=c_0$, so $X^*=\ell^1$. Define $T:\ell^1\to\ell^1$ by $$Tx=T(x_1,\dots)=\left(\sum_{j=1}^\infty x_j,0,0,0,\dots\right).$$
Let $V=\{x\in\ell^1:|x_1|<1\}$. If you go back to the definitions you can verify that $V$ is a weak* neighborhood of the origin but there is no weak* neighborhood of the origin mapped into $V$ by $T$. (Given $\delta>0$ and $y^1,\dots,y^n\in c_0$, there exists $x\in\ell^1$ such that $\left|\sum_jx_jy^k_j\right|<\delta$ for $k=1,\dots, n$ but $Tx\notin V$.)
What's Really Going On Here: If $T:X^*\to X^*$ is weak*-weak* continuous it turns out that the adjoint $T^*$ must map $X$ into $X$. But here $$T^*(1,0,0,0,\dots)=(1,1,1,1,\dots),$$so $T^*(c_0)\not\subset c_0$.
We can consider the following more general question. Suppose $X$ and $Y$ are normed spaces and $T\colon X^{*}\to Y^{*}$ is a norm-norm continuous linear operator. Does it follow that $T$ is weak$^{*}$-weak$^{*}$ continuous?
In general, the answer is no. For a counterexample, let $X$ be any non-reflexive normed space and $Y$ be any non-zero normed space. Since $X$ is not reflexive there is some $F\in X^{**}$ which is not weak$^{*}$ continuous. Take $g\in Y^{*}\setminus \{0\}$ and define $T\colon X^{*}\to Y^{*}$ by $T(f) := F(f) g$. Since $F\in X^{**}$ we have that $T$ is a norm-norm continuous linear operator from $X^{*}$ to $Y^{*}$. However, we also have ${\rm ker} \, T = {\rm ker} \, F$ and ${\rm ker} \, F$ is not weak$^{*}$ closed in $X^{*}$. Hence ${\rm ker} \, T$ is not weak$^{*}$ closed in $X^{*}$. This shows that $T$ is not weak$^{*}$-weak$^{*}$ continuous.
However, if $X$ is reflexive then the answer is yes. Because $T\colon (X^{*}, {\rm norm}) \to (Y^{*}, {\rm norm})$ is continuous we deduce that $T\colon (X^{*}, {\rm weak}) \to (Y^{*}, {\rm weak})$ is continuous. Then as the weak and weak$^{*}$ topologies on $X^{*}$ coincide on the reflexive space $X$ it follows that $T\colon (X^{*}, {\rm weak}^{*}) \to (Y^{*}, {\rm weak})$ is continuous. Finally, since the weak$^{*}$ topology is coarser than the weak topology, the identity map ${\rm id}\colon (Y^{*}, {\rm weak}) \to (Y^{*}, {\rm weak}^{*})$ is continuous, so by composing the two maps we obtain that $T\colon (X^{*}, {\rm weak}^{*}) \to (Y^{*}, {\rm weak}^{*})$ is continuous.
There is also a characterisation of weak$^{*}$-weak$^{*}$ continuity of such operators. If $X$ and $Y$ are normed spaces and $T\colon X^{*}\to Y^{*}$ is linear, then $T$ is weak$^{*}$-to-weak$^{*}$ continuous if and only if there is some norm-to-norm continuous linear $S\colon Y\to X$ such that $T = S^{*}$. See here for more details.