# Bounding convex functions by hyperplanes

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.)

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 $$λ(x) + c - 1 < f(x) ≤ g(x) < λ(x) + c + 1$$