# Strong convergence does not imply operator norm convergence

How do I construct a sequence of bounded linear transformations that converge strongly to the zero operator, but do not converge to the zero operator in the operator norm? Something strange must happen for certain elements of the Hilbert space, but what sort of thing should I be looking for? An example would be most helpful, thanks in advance!

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## 2 Answers

Hint: Consider iterated powers of the unilateral left shift operator $S\colon \ell^2(\mathbb N)\to \ell^2(\mathbb N)$ given by $$(Sa)(n)=a(n+1).$$ Can you compute $\lVert S^ka\rVert$ in terms of the entries of $a$? On the other hand, what is the operator norm of $S^k$?

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sorry, what is the operator norm in this? I'm quite confused on this point. – anegligibleperson Aug 14 '12 at 18:28
@anegligibleperson The operator norm is $\|S^k\| = \sup_{\|x\|\le 1} \|S^kx\|$. – martini Aug 14 '12 at 18:33
@martini Thanks, I'm studying from Stein and Shakarchi, and the definition used seems different, I'll think about it then. – anegligibleperson Aug 14 '12 at 18:42
Sorry, I'm probably being stupid here, but why does the operator norm not converge to 0 in this case? What about using the right shift operator instead? – anegligibleperson Aug 14 '12 at 19:06
@anegligibleperson: No. In fact, for every $k\in\mathbb N$ you can find an element $a\in\ell^2(\mathbb N)$ such that $\lVert S^k a\rVert=\lVert a\rVert$. Do you see which? So the norm of each $S^k$ has to be at least $1$. It actually equals $1$. – Rasmus Aug 15 '12 at 6:09

Another option is to fix an orthonormal basis $\{e_j\}$ and consider the orthogonal projections $P_j$ given by $$P_j\xi=\langle\xi,e_j\rangle e_j.$$ Then $\|P_j\|=1$ for all $j$, and $P_j\to0$ strongly. Indeed, for any vector $\xi$ we have by Parseval $$\|\xi\|^2=\sum_j\|P_j\xi\|^2,$$ and in particular $\|P_j\xi\|\to0$.

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