# If $C$ is compact operator and $M_n$ converges pointwise, show that $CM_n$ converges uniformly.

Suppose $C$ is a compact operator and $M_n$ is a sequence of bounded linear operators converging pointwise to another bounded linear operator $M$. Show that $||CM_n - CM|| \rightarrow 0$.

I know how to prove $||M_n C - MC|| \rightarrow 0$ (using Ascoli-Arzela theorem) but get stuck after changing the order of $M_n$ and $C$. Can anyone shed some light? All hints will be appreciated!

• I think it is easier than using Ascoli-Arzela. You need to prove that $$\|CM_n-CM\|= \sup_{\|x\|\leq 1} \|CM_nx-CMx\|\to 0.$$ By contradiction: $\exists\ \varepsilon>0$ such that $\forall\ n$ there exists $N>n$ such that $\|CM_N-CM\|>\varepsilon$, which means that there is an $x_N,\|x_N\|\leq 1$ such that $\|CM_Nx_N-CMx_N\|>\varepsilon$. Therefore we can find a subsequence $x_{n_k}$, which can be immediately renamed $x_n$ for simplicity, such that $$\|CM_nx_n-CMx_n\|>\varepsilon,\ \|x_n\|\leq 1.$$ Now we should try to use compactness. – Tommaso Seneci Dec 9 '17 at 14:57

## 1 Answer

You cannot prove that from the assumptions, since $(CM_n)$ need not converge in norm.

Consider your favourite $p \in [1, +\infty)$ and let $X = \ell^p(\mathbb{N})$, $C$ the multiplication operator given by

$$(Cf)(k) = 2^{-k}f(k)$$

and $M_n$ the left shift by $n$ places,

$$(M_nf)(k) = f(k+n).$$

Then $\lVert M_n\rVert = 1$ for all $n$, and $M_nf \to 0$ for every $f \in X$. But $\lVert CM_n\rVert = \lVert C\rVert$ for all $n$, so $\lVert CM_n\rVert \not\to 0$.

• Thanks, this is a nice counter-example! – Dormire Dec 9 '17 at 18:02