Given norm-to-norm continuous map $T:\ell^{\infty}\rightarrow L^{p}$, $p\geq 1$. Does norm-to-norm continuous imply weak*-to-weak continuous?
I have learnt that: if $T$ is norm-norm continuous then it is weak-weak continuous, what about weak*-to-weak continuous?
The bot mention A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for some $T\in B(X,Y)$, I am a green hand, but I think weak* continuous is weak* -to-norm continuous, not weak*-to-weak continuous.
Write $X = l^\infty$ and $Y=L_p$. We will use: $X$ is a dual space but not reflexive, and $Y \ne \{0\}$.
Since $X$ is not reflexive, there is a bounded linear functional $\phi$ on $X$ that is not weak* continuous. Choose a nonzero vector $u \in Y$. Define rank-one operator $T : X \to Y$ by $$ T(x) = \phi(x) u,\quad x \in X . $$ Then $T$ is norm-norm continuous, but not weak*-weak continuous.
If $T$ is norm continuous. Assume $x_n\rightharpoonup x$ for a continuous linear form so that $f\circ T$ is also a linear continuous form. Hence $f\circ Tx_n\to f\circ Tx$. This means that $ Tx_n\rightharpoonup Tx$.
Conversely, assume $T$ is weakly continuous. If $T$ is not norm continuous, then in particular $T$ is not continuous at the origin $x=0$. Thence there is $r>0$ such that for all $n\geq 1$ there is $x_n$ such that $\|y_n\|\leq \frac{1}{n^2}$ and $\|Ty_n\|\geq r$
Letting $x_n= ny_n$ we find that $\|x_n\|\leq\frac1n\to 0$ and $\|Tx_n\|= n\|Ty_n\|\geq rn\to\infty.$
In particular $x_n\rightharpoonup 0$ weakly and hence the weak continuity of $T$ implies $Tx_n\rightharpoonup 0$ weakly, which in turn implies that $(Tx_n)_n$ is norm bounded. This is clearly a contradiction since $\|Tx_n\|\to\infty.$ Necessarily $T$ is norm continuous.
