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Let $K$ be a compact Hausdorff space and denote by $C(K)$ the space of all real valued and continuous functions on $K$. We endow $C(K)$ with the supremum norm topology, making it a Banach space.

Suppose now that $V \subseteq C(K)$ is a linear subspace with the following property: For every open subset $U \subseteq K$ and every $x \in U$ there exists $f \in V$ with $f(K) \subseteq [0,1]$ such that $f(x) = 1$ and $f = 0$ outside of $U$.

Does it follow that $V$ is dense in $C(K)$?

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    $\begingroup$ One may ask a further question: what is the maximum co-dimension of closed linear subspaces $V\subset C(K)$ with the stated property? $\endgroup$
    – Pietro Majer
    Commented Nov 1, 2023 at 12:07

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Here is a counterexample for $K:=[0,1]$. There exists a Borel set $B\subset K$ that meets every non-empty open set $A\subset K$ in a subset of it of positive, not full Lebesgue measure: $0<|A\cap B|<|A|$. It’s a classic exercise by Rudin (Real and Complex Analysis, Chap.2).

Consider the bounded linear form on $C(K)$ $$\psi(f):=\int_Bf(t)dt-\int_{K\setminus B}f(t)dt.$$ I claim the closed hyperplane $V:=\ker\psi$ of $C(K)$ has the stated property. In order to prove it, let's define functions $$g_{x,m}(t):=(1-m|x-t|)_+$$ for $x\in K$ and $m>0$. Note that $|\psi(g_{x,m})|\le\frac1m$, and if $B$ has density $0$ resp. $1$ at $y$, then $\psi(g_{y,m})=-\frac1m(1+o(1))$ resp. $\psi(g_{y,m})=\frac1m(1+o(1))$, as $m\to+\infty$. Then the function $$f :=g_{x,2m}+\lambda_m g_{y,m},$$ $$ \lambda_m:=-\frac{\psi(g_{x,2m})}{\psi(g_{y,m})} $$ for suitable $m$ and $y$ verifies $$0\le f\le 1,\qquad f^{-1}(1)=\{x\},\qquad \text{supp}(f)\subset U,\qquad \psi(f)=0,$$ proving the claim. In order to choose $m$ and $y$, we first fix points $y_0$ and $y_1$ in $U\setminus\{x\}$ of density $0$ resp. $1$ for $B$, which implies $\limsup_{m\to\infty}|\lambda_m|<1$, both for the choice $y:=y_0$ and $y:=y_1$. Then we fix $m$ so large that $|\lambda_m|<1$ and that $g_{x,2m}$, $g_{y_0,m}$ and $g_{y_1,m}$ have their supports in $U$, pairwise disjoint, and lastly we choose $y\in\{y_0,y_1\}$ so that $\lambda_m\ge0$.

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    $\begingroup$ Can you explain a) how to choose $g$? b) how to fix $m, \epsilon$? $\endgroup$
    – Giorgio Metafune
    Commented Oct 31, 2023 at 22:44
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    $\begingroup$ a) The function $g$ can be taken as well of the form $g(t):=(1-n|y-t|)_+$, with $y\in U\setminus\{x\}$ a Lebesgue point of either $B$ of $K\setminus B$. THis make the sign of $\psi(g)$ positive resp. negative, if $n$ is large enough; one also takes $n$ so large that $g$ has support in $U\setminus\{x\}$. Say we do both: a $g_-$ and a $g_+$, with $\psi(g_-)<0<\psi(g_+)$ $\endgroup$
    – Pietro Majer
    Commented Oct 31, 2023 at 23:02
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    $\begingroup$ b) Then one fixes $m$ so large that $(1-m|x-t|)_+$ has support in $U$, disjoint from the supports of $g_+$ and $g_-$, and such that $\psi(g_-)<\psi((1-m|x-t|)_+)<\psi(g_+)$ (note that $\psi((1-m|x-t|)_+)=o(1)$ as $m\to\infty$). Finally one chooses $g_-$ or $g_+$, and fices $0\le\epsilon<1$ $\endgroup$
    – Pietro Majer
    Commented Oct 31, 2023 at 23:12
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    $\begingroup$ And what about the counterexample with the discrete measure you added? I had a look and seemed to work? $\endgroup$
    – Giorgio Metafune
    Commented Nov 1, 2023 at 12:12
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    $\begingroup$ Actually I see a (minor?) problem with the version with the discrete measure. The functional is now $$\Psi:=\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\delta_{q_k}$$ for an enumeration $\{q_k\}_k$ of $\mathbb Q\cap [0,1]$. Then for all $k$ we have $\Psi(g_{q_k,n})=\frac{(-1)^k}{k^2}+o(1)$ as $n\to+\infty$. So the same construction with $f=g_{x,m}+\lambda g_{y,n}$ works, but for the case where $x=q_k$, for odd $k$: I do not see how to make $\Psi(f)\ge0$ and $f\ge0$ then. $\endgroup$
    – Pietro Majer
    Commented Nov 1, 2023 at 13:14

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