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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

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See also… – user16299 Jan 13 '12 at 20:09
up vote 11 down vote accepted

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 $Y$.

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Uniform boundness principle requires that the space $X$ is complete, but in the original question this wasn't assumed. – Norbert Jan 13 '12 at 19:03
@Norbert I added some additional detail. Perhaps I was too cavalier with my phrasing. – David Mitra Jan 13 '12 at 19:15

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