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Suppose $f : [0, 1]^\alpha \to \mathbb{R}$ is convex. Does it follow that $f$ has a lower bound?

I've shown that this holds for finite $α$. My question is whether it also holds for infinite $α$.

In the finite case, I can prove it by induction. First, we show this for an interval $[0, 1]$. (For convex $f : [0, 1] \to \mathbb{R}$, for $x ≥ \frac{1}{2}$ we have $f(x) ≥ 2f(\frac{1}{2}) - f(0)$, and a similar bound holds for $x ≤ \frac{1}{2}$.)

Next, suppose that $f : [0, 1]^{n + 1} \to \mathbb{R}$ is convex. Then for each $x ∈ [0, 1]$, let $f_x$ be the function that takes $y ∈ [0, 1]^n$ to $f(x, y_1, …, y_n)$. Then each function $f_x$ is convex, and thus by the inductive hypothesis it is bounded. Let $b(x)$ be the greatest lower bound of $f_x$. Then $b : [0, 1] \to \mathbb{R}$ is also convex. (For any $ε > 0$, there are $y$ and $y'$ such that $f_x(y) < b(x) + \frac{ε}{2}$ and $f_{x'}(y') < b(x') + \frac{ε}{2}$. So \begin{align} b(λx + (1-λ)x') & ≤ f(λx + (1-λ)x', y) \\ & ≤ λf(x, y) + (1 - λ)f(x', y') \\ & < λb(x) + (1 - λ)b(x') + ε \end{align} So $b(λx + (1-λ)x') ≤ λb(x) + (1 - λ)b(x')$.) So by the first part, $b$ also has a lower bound, and this must be a lower bound for $f$.

It seems like a similar argument using transfinite induction might work in general, but I don't know how the limit step would go.

Jeff Russell
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  • Have you tried something like an $f$ such that $f(x) = 0$ is $0$ if $x_i$ is non-zero for infinitely many $i$ and is $\sum_i x_i$ otherwise? – Rob Arthan Oct 02 '18 at 23:35
  • Nope, I didn't think of that. Looks like that works as a counterexample (if we just change the sign). – Jeff Russell Oct 03 '18 at 00:51
  • Oh, wait, I was mixed up. If we keep the sign the same, we still have a lower bound. If we switch the sign, though, this is no longer convex, but rather concave. Maybe I'm missing something, but I'm not seeing how to get a counterexample out of this. – Jeff Russell Oct 03 '18 at 03:24

1 Answers1

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No, it may not be bounded below. Consider $K=[0,1]^\alpha$ as a subset of the Banach space $\ell^\infty(\alpha)$. $K$ is a neighbourhood of the point $u=\left(\frac12,\frac12,\cdots\right)$. If $\alpha\ge\aleph_0$, then there is a discontinuous linear functional $\phi:\ell^\infty(\alpha)\to \Bbb R$. Discontinuous functionals are unbounded (in both directions) on every open sets, therefore $f(x)=\phi(x-u)$ is a convex function $K\to\Bbb R$ without lower bounds.

  • Cool, very helpful. Now that I know what terms to search for, it looks like this answer is relevant. The answer provides a more explicit construction of such a function. Let $α = ω$. We can extend the set of vectors $e_n$ to a basis for $\mathbb{R}^ω$. Then there is a unique linear function $f : \mathbb{R}^ω \to \mathbb{R}$ such that $f(e_n) = n$, and $f(v) = 0$ for the other basis vectors $v$ besides the vectors $e_n$. It's clear that $f$ is unbounded. – Jeff Russell Oct 03 '18 at 03:40