Continuity in weak convergence implies continuity in norm convergence Let $X$ and $Y$ be norm spaces, and let $T:X\to Y$ be a linear transformation which is continuous under weak convergence.
That is, if $\forall x^\star\in X^\star:x^\star x_n\to x^\star x $ then $\forall y^\star \in Y^\star:y^\star Tx_n \to y^\star Tx$.
Prove that $T$ is continuous under norm convergence.
That is, if $\|x_n\|\to\|x\|$ then $\|Tx_n\|\to\|Tx\|$.
Any hints would be most welcomed.
TIA,
Shai
 A: Suppose that  $T$ is not continuous under norm convergence. Then there is a sequence $\{x_n\}$   such that
$$ \Vert x_n\Vert\le 1,\quad \text { and }\quad\Vert T(x_n)\Vert \ge n^2 \quad\text{ for each }n.  

$$
(there is a sequence $z_n$ of non-zero vectors in $X$ converging in norm to $0$ for which $\lim\limits_{n\rightarrow\infty} \Vert Tz_n\Vert\ne 0$. Then $\{z_n/\Vert z_n\Vert\}$ is in the unit ball of $X$ and  $\{T( z_n/\Vert z_n)\Vert\}$ is unbounded).
Now $\Vert x_n/n \Vert\rightarrow 0$, and thus $y^* (x_n/n)\rightarrow0$ for each $y^*\in Y^*$.  Since $T$ is assumed to be continuous under weak convergence, $$y^*T(x_n/n)\rightarrow 0\quad, \text{ for any }y^*\in Y^*.   $$
But then,  it follows from the  Uniform Boundedness Principle$^\dagger$ that $\{T(x_n/n)\}$ is norm bounded. 
However, $\Vert T(x_n/n)\Vert\ge n$ for each $n$.

$^\dagger$ We consider $\{Tx_n :n=1,2,\ldots\}$ as a subset of $Y^{**}$. That is we consider $Tx_n$ as a continuous linear functional on $Y^*$ (which is a Banach space). The $Tx_n$  are  pointwise bounded, and, thus, norm bounded in $Y^{**}$. But the norms of the $Tx_n$ in $Y^{**}$ are the same as the norms in $Y$.
