Let $X$ be a compact metric space, and let $C(X)$ be the Banach space of continuous real-valued function on $X$, with the maximum norm.

Suppose $S\subset C(X)$ is a set of functions with the following property:

For every ball $B(a,r)\subset X$ and for every $\epsilon>0$, there exists a function $f\in S$ such that:

(i) $0\leq f(x)\leq 1$ for all $x\in X$,

(ii) $f(a)=1$,

(iii) $|f(x)|<\epsilon$ for $x$ outside the ball $B(a,r)$.

My question: does the above assumption on $S$ imply that the set $S$ spans $C(X)$, that is that every continuous function on $X$ can be arbitrarily approximated (in the max norm) by finite linear combinations of functions in $S$?

(An answer in the special case $X=[0,1]$ would also be of interest to me).

  • $\begingroup$ No. For $X=[0,1]$, you can let $S$ be the set of functions with $\int\_0^1f(x)\,dx=0$. $\endgroup$ – George Lowther Sep 20 '11 at 20:42
  • 1
    $\begingroup$ Thanks George... I have corrected the question to demand that the functions in $S$ be positive. This is the situation I am really interested in. $\endgroup$ – user17970 Sep 20 '11 at 20:49
  • $\begingroup$ ... and bounded from above by $1$ $\endgroup$ – user17970 Sep 20 '11 at 20:55
  • 1
    $\begingroup$ Still no. Let $\mu$ be any finite signed measure on $[0,1]$ whose positive and negative parts $\mu^+,\mu^-$ have full support (nonzero measure on every nonempty open set). Then let $S\subseteq C([0,1])$ be the functions $f$ with $\int f\,d\mu=0$. $\endgroup$ – George Lowther Sep 20 '11 at 20:57
  • 1
    $\begingroup$ ...ok, bounded above by one. Still no, with the example I just gave (the signed measure just has to have no atoms for $S$ to satisfy your property). $\endgroup$ – George Lowther Sep 20 '11 at 20:59

No, $S$ does not have to span $C(X)$.

Taking the case with $X=[0,1]$, let $\mu$ be any atomless finite signed measure whose positive and negative parts $\mu^+$,$\mu^-$ have full support, so that $\mu^+(U) > 0$ and $\mu^-(U) > 0$ for any nonempty open $U$. Then, the set $S=\{f\in C(X)\colon\int f\,d\mu=0\}$ satisfies your properties, is closed in $C(X)$, but is not all of $C(X)$.

To see that $S$ satisfies your properties, consider any $a\in U$ for $U$ an open subset of $X$. Then choose $r > 0$ such that $V=U\setminus \bar B_r(a)$ is nonempty. There exists a nonnegative $g\in C(X)$ with support in $V$ such that $\mu(g) > 0$. Otherwise we would have $\mu^+(S)\le\mu^-(S)$ for all Borel $S\subseteq V$, which would imply that $\mu^+(V)=0$ contradicting the assumption that $\mu^+$ has full support. Similarly, there is a nonnegative $h\in C(X)$ with support in $V$ and $\mu(h) < 0$. By scaling, we can assume that $g,h$ are bounded by $1$.

Now, the functions $f_n(x)=\max(1-n\vert x-a\vert,0)$ have support in $B_r(a)$ (for $n > 1/r$) and decrease to $1_{\{x=a\}}$ as $n\to\infty$. As $\mu$ is atomless, we have $\mu(f_n)\to0$. Choosing $n$ large enough that $\mu(h) < \mu(f_n) < \mu(g)$ then $f=f_n+\lambda h$ or $f=f_n+\lambda g$ will satisfy $\mu(f)=0$ for some $0\le\lambda\le1$. But, $f\in S$ satisfies (i) $0\le f\le1$, (ii) $f(a)=1$, and (iii) $f=0$ outside of $U$.

As such measures $\mu$ will always exist on any compact metric space without isolated points, your conclusion does not hold on any compact metric space other than when $X$ is countable.

  • $\begingroup$ Can you clarify one point: why would $\mu^+(S)\leq \mu^-(S)$ for all Borel $S\subset V$ imply that $\mu^+(V)=0$? I think you have to posit in advance that the measure $\mu$ is not positive and not negative on any open set. $\endgroup$ – user17970 Sep 21 '11 at 7:02
  • $\begingroup$ .. I mean: to posit that the restriction of $\mu$ to any open set is not a positive measure and not a negative measure. $\endgroup$ – user17970 Sep 21 '11 at 7:12
  • $\begingroup$ @guykatriel: By the Hahn-Jordan decomposition you have a Borel set $A$ with $\mu^+(E)=\mu(E\cap A)$ and $\mu^-(E)=-\mu(E\setminus A)$ for all $E$. If $\mu^+(S)\le\mu^-(S)$ for all Borel $S\subseteq V$ then that would imply $$0\le\mu^+(V)=\mu^+(V\cap A)\le\mu^-(V\cap A)=0.$$ $\endgroup$ – George Lowther Sep 21 '11 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.