Suppose that you have a bounded function $f(x)$ on a compact domain in $\mathbb{R}^n$. It's easy to see from Holder's inequality that $$ ||f||_1 \leq \operatorname{Volume}(D) ||f||_\infty. $$ There can't be a reverse inequality for arbitrary functions by the usual "tall, thin spike" counterexamples.

But suppose that we know, in addition, that the derivative of $f$ is uniformly bounded by some K and the domain $\Omega$ is nicely shaped (say, a disk). Then I think there is a reverse inequality in the form $$ ||f||_\infty \leq C(K,\Omega) ||f||_1 $$ where $C$ depends on the domain and the derivative bound. I'd actually like to know $C(K,\Omega)$ for the unit disk, or at least a bound on it.

I'm sure that this must be a standard result, but I don't know the name of it. Does anyone know where this is written down? (Or which standard family of inequalities it follows from?)

Plausibility argument: An informal argument would be that we can assume $||f||_\infty = f(x) > 0$ at some $x$, and that because the function has bounded derivative, there is a cone of height $f(x)$ (and slope given by the derivative bound) beneath the graph of $|f|$. The volume of this cone should be proportional to $f(x)$ and bound $||f||_1$ below (assuming the whole cone is inside the disk).

However, this argument seems to be a little ugly in practice; if $x$ is close to the boundary or $f(x)$ is very large, the entire base of the cone might not be contained in the disk, for instance, so there seem to be various cases to work through.

Notice also that this argument won't work (and I think the result isn't true) on an arbitrary compact domain, so somehow the shape of the domain has to be part of the argument; long, thin, ``tendrils'' would allow even a function of bounded derivative to achieve a large value without contributing much to the integral.