# Different types of continuity for operators on Hilbert spaces

In chapter one of K-theory and $C^*$-algebras, a Friendly Approach, the author gives a very brief discussion about several types of continuity of operators between Hilbert spaces.

Let $T: \mathcal{H}_1\to\mathcal{H}_2$ be a linear operator between Hilbert spaces, we say $T$ is $\tau_1-\tau_2$ continuous if $T$ is continuous when $\mathcal{H}_1$ is equipped with topology $\tau_1$ and $\mathcal{H}_2$ is equipped with $\tau_2$.

The author says $T$ is norm-norm continuous iff $T$ is norm-weak continuous iff $T$ is weak-weak continuous, and that $T$ is weak-norm continuous iff $T$ is of finite rank.

I know that norm-norm countinuous implies norm-weak continuous, and weak-weak continuous implies norm-weak continuous. But I cannot figure out the rest of the assertion.

Can somebody give a hint? Also, before I continue to read the later parts of the book, I wonder whether these different types of continuity plays a crucial part in the theory.

Thanks!

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A variant can be obtained as follows: let $B$ be the closed unit ball in $\mathcal{H}_1$, and equip it with the weak topology. Then $T$ is a compact operator if and only if $T|_B\colon B \to \mathcal{H}_2$ is continuous (with respect to the norm topology on $\mathcal{H}_2$). See here. –  t.b. Jul 13 '12 at 19:20

If $T$ is norm-weak continuous, apply the Principle of Uniform Boundedness to the family of linear functionals $x \to \langle y, Tx \rangle$ for $\|y\| \le 1$ to conclude that they are uniformly bounded, and therefore that $T$ is a bounded linear operator (i.e. norm-norm continuous).
To show $T$ is weak-weak continuous, note that $\langle y, Tx \rangle = \langle T^* y, x \rangle$.
If $T$ is weak-norm continuous, there is a finite set $\{y_1, \ldots, y_n\}$ such that $\|Tx\| \le 1$ whenever all $|\langle y_j, Tx \rangle| \le 1$. In particular, if all $\langle y_j, Tx\rangle = 0$ we must have $Tx=0$. But in any vector space of dimension $>n$, the intersection of $n$ hyperplanes contains a nonzero vector. So $\text{Ran}(T)$ has dimension at most $n$.