Let $(X,d)$ be a separable and connected metric space. My question is rather short and to the point: do there exist $\{x_n\}_{n=0}^{\infty}\subseteq X$ such that $$ \left\{d(x_n,\cdot)-d(x_0,\cdot)\right\}_{n=1}^{\infty}, $$ is a Schauder basis of $C_b(X)$? If so, what is this "basis" called in the literature?
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$\begingroup$ Do you know any example where $C_b(X)$ is separable, yet does not admit a Schauder basis? $\endgroup$– Gerald EdgarCommented Jan 28, 2022 at 9:56
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1$\begingroup$ Two remarks which might be of interest. 1. A standard reference on your topic is the Springer Lecture Notes "Schauder Bases in Banach Spaces of Continuous Functions" by Z. Semadeni. 2. As a general rule of thumb, if you want to extend results and concepts on spaces of bounded, continuous functions on compact spaces to the non-compact case (Gelfand-Naimark, duality for bounded Radon measures, tensor product representations,...), one way to go is to replace the norm by the strict topology (R.C. Buck), i.e., the finest l.c. topology which agrees with compact convergence on the unit ball. $\endgroup$– bathalf15320Commented Jan 28, 2022 at 10:32
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1$\begingroup$ @Carl_Petterson Take any element $f$ of $C_B(\mathbb{N})$. For each $n$, let $f(n\pm 1/2)=0$ and connect the values of $f(n\pm 1/2)$ and $f(n)$ by a straight line. $\endgroup$– Michael GreineckerCommented Jan 28, 2022 at 10:33
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1$\begingroup$ I only know of an existence proof using the Pełczyński decomposition method so it won't give an explicit basis. For these, I would try Semadeni. $\endgroup$– bathalf15320Commented Jan 28, 2022 at 13:45
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1$\begingroup$ @Carl_Petterson, this is far from isometry, only isomorphism. Otherwise it would have contradicted the Banach-Stone theorem. $\endgroup$– Tomasz KaniaCommented Jan 28, 2022 at 14:14
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Note that in order for $C_b(X)$ to have a Schauder basis, $X$ has to be compact. Indeed, $C_b(X)$ is naturally isomorphic to $C(\beta X)$ and the latter is non-separable (because $\beta X$ is non-metrisable) as long as $X$ is a non-compact metric space.
Thus, you are in the realm of compact metric space to have a chance for a Schauder basis of some specific form since $C(X)$ always has a Schauder basis for a compact metric connected space. But do you have a basis of the required form already for $C[0,1]$?
answered Jan 28, 2022 at 10:14
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$\begingroup$ @TomaszKanie I was hopeing to use this: arxiv.org/abs/1305.1529 $\endgroup$ Commented Jan 28, 2022 at 10:56