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This is a follow-up of a question of mine with a similar title. I am interested in Morse homology (on Hilbert manifolds), more specifically with "generic" perturbations of the metric tensor (under the heading of "transversality"). The space of perturbations to use should have the property of separability in order to apply the Sard-Smale theorem.

Now comes the question, which is supposed to help me in these matters: Let H be a separable Hilbert space and let B be the closed unit ball in H. Is the space C^b(B), the space of continuous bounded functions on the closed unit ball endowed with the sup-norm, a separable space?

The previous question of mine replaced B above with an open subset of the Hilbert space. Then the answer turns out to be NO. Notice that if H is finite dimensional, the answer is YES.

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It is not separable. Counterexamples are obtained by putting translates of a suitable bump function at orthogonal half-unit vectors and summing them in $2^\infty$−many ways. More explicitly, consider all $f=\sum_{i\in\mathbb N}(\lambda_i\phi_i)$ with $\lambda_i\in\{0,1\}$ . –  TaQ Feb 2 '11 at 17:56
    
$B$ is homeomorphic to $H$ [due to Klee]. $H$ is homeomorphic to $\ell^{1}$ [due to Kadec]. And $\ell^{1}$obviously is homeomorphic to its open unit ball $U$. Now, $\ell^{\infty}$, viewed as a dual, clearly is isometrically isomorphic to a [complemented] subspace of $C_{b}\left(U\right)$ (simply take the restrictions of the functionals). –  Ady Feb 3 '11 at 14:27
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up vote 6 down vote accepted

No. Ady's construction still works, I think. Here's another, easier one:

Choose an orthonormal basis $\{e_{n}\}$ of $H$. Since $\|e_{i} - e_{j}\| = \sqrt{2}$, the continuous functions $f_{n}(x) = \max\{0, 1 - 2 \cdot \|e_{n} - x\|\}$ have disjoint support and sup-norm $1$. This gives an isometric embedding $\ell^{\infty} \to C_{b}(B)$ by sending a bounded sequence $(a_{n})$ to $\sum a_{n} f_{n}$.

A simple modification of this argument shows that $C_{b}(X)$ contains a copy of $\ell^{\infty}$ (and thus isn't separable) whenever the completely regular space $X$ has a countable discrete subset.

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All but one of the functions $\sum a_nf_n$ of Theo Buehler have "teeth" on the boundary of $B$ . This is not a flaw here, but if for some reason one wished smooth functions which are zero on the boundary, one might prefer something like $f_n(x)=\exp\,\big(\,1+(\,9\,\|\,x-\frac 12\,e_n\|^{\,2}-1){}^{-1}\big)$ for $\|\,x-\frac 12\,e_n\|<\frac13$ , and $f_n(x)=0$ otherwise. Here $x\mapsto\|x\|$ must be the inner product norm. –  TaQ Feb 3 '11 at 13:25
    
@try-a-question: thanks! –  Theo Buehler Feb 3 '11 at 13:53
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