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 $T^*(c_0)\not\subset c_0$.

  • $\begingroup$ Can yo tell how $T^*(1,0,0,0,0...)=(1,1,1,1,1...)?$ $\endgroup$ – user510271 Feb 11 '19 at 6:46
  • $\begingroup$ @user510271 Do you know the definition of $T^*$? $\endgroup$ – David C. Ullrich Feb 11 '19 at 13:07

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