Is every convex function the supremum of a countable family of affine functions?

Question

Let $V$ be a separable normed space and $M$ an open convex subset of $V.$ Then $M$ has a countable dense subset $D.$ If $f$ is convex, then at every point $t$ of $D$ there is a support, namely an affine function $A$ such that $A \leq f$ on $M$ and $A(t) = f(t)$. Assume $D = (t_n)$ is an enumeration of $D$ and choose $A_n$ any support of $f$ at $t_n.$ Let $\varphi = \sup A_n,$ which is convex satisfying that $\varphi \leq f$ on $M$ and $\varphi = f$ on $D.$

Now, if $f$ were continuous (which is guaranteed if $\dim V < \infty$), then so would be $\varphi$ and by density of $D,$ we would have $f = \varphi$ on $M$ (not just on $D$).

Is continuity of $f$ essential in concluding $f = \varphi$? By that I mean, can any convex function $f:M \to \mathbf{R}$ be written as the supremum of a countable family of affine functions?

Answer 1

This is not a complete (pun intended) answer, but too long for a comment.

If the space $V$ is complete, then every convex and lower semi-continuous function $f \colon M \to \mathbb R$ is automatically continuous. Hence, you need an incomplete space or a non-lsc function $f$.

An interesting situation is the following. Let $c_c$ be the space of finite sequences equipped with the $\ell^2$-norm. Then the function $f \colon c_c \to \mathbb R$, $$ f(x) = \sum_{n=1}^\infty n x_n^2, $$ is convex, lower semicontinuous, but nowhere continuous. Note, however, that this function possesses a continuous support line at every point (which contradicts your claim in a comment).

I did not checked the details, but I could imagine that we could choose a dense subset $D$ consisting of points of the form $$ d := \hat x + \varepsilon e_n $$ where $\hat x$ is a rational approximation on the first $m$ components, $\varepsilon > 0$ is small and $n \in \mathbb N$ is very large. Now, given any $x \in c_c$, we find a $d \in D$ like above such that $x - \hat x$ is very small. If take the supporting function at $\hat x$ and evaluate it at $x$, we get $$ A_{\hat x}(x) = f(\hat x) + f'(\hat x)(x - \hat x) \approx f(x) + n \varepsilon^2 - 2 n \varepsilon^2. $$ This is much smaller than $f(x)$ (if, in the construction of $D$, we have taken care such that $n$ is really very large).

Answer 2

We endow $V'$ with the weak-star topology. And we assume that $V'$ is separable with respect to this topology. If $V$ is separable then $V'$ seems to be weak-star separable. (Need to look for references).

Claim: Let $V$ be a normed space with $V'$ weak-star separable. Let $f: V \to \mathbb R \cup \{+\infty\}$ be proper, convex and lower semicontinuous. Then $f$ is the supremum of countably many affine and continuous functions.

Then there is an affine and continuous function $g$ such that $g \le f$.

Now let $$ A := \{ (g,a) \in V' \times \mathbb R: \ g(x) + a \le f(x) \quad \forall x\in V\}. $$ This set is non-empty. Then $$ f(x) = \sup_{ (g,a) \in A} g(x) + a . $$ (These two claims need Hahn-Banach, but do not need completeness of the space.)

Since $V'$ is separable, also $V' \times \mathbb R$ is separable. Let $B \subseteq A$ a countable and dense subset. Define $$ \phi(x):= \sup_{ (g,a) \in B} g(x) + a . $$ Clearly $\phi \le f$.

Take $x\in V$ and $\epsilon>0$. Then there is $(g,a) \in A$ such that $$ f(x) \le g(x) + a + \epsilon. $$ Due to the weak-star separability, there is $(h,b) \in B$ such that $ (g-h)(x) < \frac\epsilon 2$ and $|a-b|<\frac\epsilon 2$. Then $$ \begin{split} f(x) &\le g(x) + a + \epsilon \\ &\le h(x) + b + (g(x) - h(x)) + (a-b) + \epsilon\\ &\le h(x) + b + 2 \epsilon \le \phi(x) + 2\epsilon. \end{split}$$ Since $\epsilon$ was arbitrary, this proves $f\le \phi$ and $f=\phi$.

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Here is a counterexample in the non-separable case. Let $V = \ell^2(\mathbb R)$, i.e., the set of functions $x:\mathbb R\to\mathbb R$ such that $x(i)\ne0$ only for at most countably many $i$, and $\sum_{i\in \mathbb R} x(i)^2 < \infty$. With the norm $\|x\|_{\ell^2(\mathbb R)} := \left( \sum_{i\in \mathbb R} x(i)^2 \right)^{1/2}$ this is a non-separable Hilbert space. We identify $V =V'$.

Now define the convex and continuous function $$ f(u):= \sum_{i\in \mathbb R} x(i)^2 . $$ Let $$ \phi(x):= \sup_{ (g,a) \in B} \langle g, x\rangle + a $$ as above with $B \subset V \times \mathbb R$ countable such that for all $(g,a)\in B$ we have $\langle g, x\rangle + a \le f(x)$ for all $x\in V$. Note that $x=0$ implies $a\le 0$ for $(g,a) \in B$. Let $$ I := \{ i \in \mathbb R: \ \exists (g,a) \in B : g(i) \ne 0\}. $$ Such $g\in V$ is only non-zero on a countable set, hence $I$ is countable. Then there is $j\in \mathbb R\setminus I$ such that $g(j) =0 $ for all $g$ with $(g,a)\in B$.

Define $x = \chi_{ \{j\}} \in \ell^2(\mathbb R)$. Then $f(x) =1$ and $\phi(x) = \sup_{(g,a)\in B}a \le 0$.