Consider a class of (proper closed) convex function on $[0,1]^d$, which we shall denote $\mathcal{F}$. If every element of $\mathcal{F}$ is bounded in $L_2$, say $$\int_{[0,1]^d} |f(x)|^2\ dx\leq 1,$$ then can we say something about the uniform lower bound of $\mathcal{F}$?
Progress
I proved the existence of a lower bound by contradiction but did not get an explicit bound.
Apparently we cannot expect a uniform upper bound.
Let $U=\{x: f(x)\le 2\}$. This is a convex set. On the complement of $U$ we have $|f|^2> 4$, hence $|U^c|\le 1/4$ (writing $|\cdot|$ for the Lebesgue measure). This gives $|U|\ge 3/4$.
Suppose $f(x_0)=-M$ for some $x_0$, where $M>2$. The set $V = \frac12x_0+\frac12U$ is contained in $[0,1]^d$ and has measure at least $3\cdot 2^{-d-2}$. By convexity, $f<1-M/2$ on $V$, hence $$ \int_{[0,1]^n}|f|^2 \ge (M/2-1)^2 |V| \ge 3(M-2)^2 2^{-d-4} $$ Consequently, $M\le 2 + \sqrt{3}\cdot 2^{d/2+2}$ which means $\inf f\ge -2- \sqrt{3}\cdot 2^{d/2+2}$.