Bounding convex functions by hyperplanes

Question

My intuition is that if a convex function and a concave function are close together, then they both must be close to flat. Is this intuition right, in infinite dimensional spaces?

More precisely, let $K$ be a non-empty convex subset of $[-1, 1]^ω$, let $f : K \to \mathbb{R}$ be a convex function, and let $g : K \to \mathbb{R}$ be a concave function. Suppose that $0 ≤ g - f ≤ 1$. I want to know whether there exists a linear function $λ : \mathbb{R}^ω \to \mathbb{R}$ and $a, b ∈ \mathbb{R}$ such that, for all $x ∈ K$, $$ λ(x) + a \; \leq f \; \leq g \; \leq \; λ(x) + b $$

(This is kind of a follow-up to this question.)

Answer 1

I think I figured it out (with help from a "similar question" in the sidebar). The answer is yes, as long as $K$ has an internal point.

The Separating Hyperplane Theorem says that if $A$ and $B$ are disjoint convex subsets of a vector space, and at least one of them has an internal point, then they can be properly separated by a nonzero linear functional.

Let $$\begin{aligned} A & = \{ (x, y) ∣ x ∈ K, \ y ≥ f(x) + 1 \} \\ B & = \{ (x, y) ∣ x ∈ K, \ y ≤ g(x) - 1 \} \end{aligned}$$ Then $A$ and $B$ are disjoint convex subsets of $\mathbb{R}^ω × \mathbb{R}$. Furthermore, $A$ has an internal point. (If $x$ is an internal point of $K$, then $(x, f(x) + 2)$ is an internal point of $A$.) So the theorem applies.

Let $h : \mathbb{R}^ω × \mathbb{R} \to \mathbb{R}$ be a nonzero linear functional that separates $A$ and $B$: for some $α$, $h(A) < α < h(B)$. We can rewrite this with a little algebra: there is a linear function $λ : \mathbb{R}^ω \to \mathbb{R}$ and a number $c ∈ \mathbb{R}$ such that for $(x, y) ∈ A$ and $(x, y') ∈ B$, $$y' < λ(x) + c < y$$ In particular, for each $x ∈ K$, $(x, f(x) + 1) ∈ A$ and $(x, g(x) - 1) ∈ B$, so we have $$ λ(x) + c - 1 < f(x) ≤ g(x) < λ(x) + c + 1 $$