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Assume $X$ is a separable Banach space with norm $|| \cdot||$. Consider $\{f_n\}_{n \in N}$ a countable dense subset of $X$ and equip $B(X)$ (bounded linear operators on $X$) with the following metric: \begin{align} & d(A,B)=\sum_{n=0}^\infty 2^{-n} (1+||f_n||)^{-1} || A f_n-B f_n || \end{align}

Does this metric make $B(X)$ complete and separable?

PS: completeness seems quite easy to prove. The interesting part is separability.

PS2: below it is shown that $B_1(X)$ (the unit ball of $B(X)$ in the operator norm metric) is separable. What about the whole $B(X)$?

http://thales.doa.fmph.uniba.sk/sleziak/texty/rozne/pozn/books/kechrisexercises.pdf

user44670
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1 Answers1

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The metric describes pointwise convergence on the dense set $\lbrace f_n:n\in\mathbb N\rbrace$ which is a vector space topology (even locally convex) which is clearly coarser than the usual operator norm topology. Therefore, the metric cannot be complete because otherwise the open mapping theorem (for Frechet spaces) implies that the metric and the norm give the same topology which is clearly not the case if $X$ is infinite dimensional.

Just an idea concerning separability: If the dual $X'$ is separable one could try to show that $X' \otimes X = \lbrace$finite rank operators$\rbrace$ is dense which would yield separability.

Jochen
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  • Hi Jochen. The metric is complete, and the proof is standard, see: http://alanmath.wordpress.com/2011/05/26/product-of-a-sequence-of-completely-metrizable-spaces-is-completely-metrizable-p13-c/ – user44670 Mar 14 '14 at 21:25
  • The only little difference with the proof I gave you is that when you take a cauchy sequence, you will get the limit operator only on the dense set ${f_n}$. Then you can extend it uniquely to a bounded operator on $X$. – user44670 Mar 14 '14 at 21:27
  • @user44670 That is a different situation. Jochen is right, $B(X)$ is not complete under the metric $d$ unless $X$ is finite-dimensional. It is clear that $p\colon A \mapsto d(A,0)$ is a seminorm. Since the sequence $(f_n)$ is dense, it is a norm. It is readily verified that $p(A) \leqslant 2\lVert A\rVert_{\text{op}}$. So $(B(X),d)$ is complete if and only if there is a $c > 0$ with $p(A) \geqslant c\lVert A\rVert_{\text{op}}$ for all $A\in B(X)$ by the open mapping theorem. But there is no such $c$ if $X$ is infinite-dimensional (which is perhaps not quite obvious). – Daniel Fischer Mar 14 '14 at 23:04
  • but you agree that this metric makes $B_1(X)$ (the unit ball of $B(X)$ in the operator norm metric) complete? – user44670 Mar 15 '14 at 20:38