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.
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.