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Let $H$ be a separable Hilbert space, show that every bounded operator from H to itself can be approximated in the strong operator topology by a sequence of finite rank operators.

I know we can find an orthonormal basis for the hilbert space. If we let $P_n$ the the projection on the first $\{e_i\}_{i=1}^n$ basis elements, how close can we come?

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up vote 2 down vote accepted

Hint: Given $T\in L(H)$, what can you say about $P_nT$?

$\def\n#1{\left\|#1\right\|}\def\p#1{\left\langle#1\right\rangle}$ Added: For each $x \in H$ we have $x = \sum_{i=1}^\infty \alpha_i e_i$ for $\alpha_i = \langle x, e_i\rangle$. We get, as the $e_i$ are orthonormal \begin{align*} \n{P_nx - x}^2 &= \n{\sum_{i=n+1}^\infty \alpha_i e_i}^2\\ &= \p{\sum_{i=n+1}^\infty \alpha_i e_i, \sum_{j=n+1}^\infty \alpha_j e_j}\\ &= \sum_{i=n+1}^\infty\sum_{j=n+1}^\infty \alpha_i \bar\alpha_j \p{e_i,e_j}\\ &= \sum_{i=n+1}^\infty |\alpha_i|^2 \end{align*} And the latter converges to 0, as $\sum_{i=1}^\infty|\alpha_i|^2 = \n x^2 < \infty$.

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I read that $\|P_nx -x\| \rightarrow 0$ but how is that? I get $\|P_nx -x \| = \|\sum_{i = n+1}^\infty \alpha_ie_i\|$ but does that need to go to 0? – Johan Dec 11 '12 at 8:43
Added sth. about it. – martini Dec 11 '12 at 8:54
Great! So this say that $P_nT \rightarrow T$ pointwise? and we are done? – Johan Dec 11 '12 at 15:51
Yes. ${}{}{}{}$ – martini Dec 11 '12 at 16:49

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