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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.

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  • $\begingroup$ Since you can approximate in $L^1$ any function by functions with bounded derivatives, there is no inequality $\|f\|_1 \leqslant C \|f\|_\infty$ with $C$ independent of $f$. However, Sobolev inequality is likely what you are looking for. $\endgroup$
    – Mateusz Kwaśnicki
    Oct 19, 2017 at 20:48
  • $\begingroup$ The title of the question is misleading, because you intend for the number $C$ to depend on the derivative bound; it's not that the $L^1$ norm and the $L^\infty$ norm on the same vector space are equivalent. $\endgroup$
    – Tom Goodwillie
    Oct 19, 2017 at 21:35
  • $\begingroup$ @Mateusz: but he allows $C$ to depend on the derivative bound. $\endgroup$
    – Nik Weaver
    Oct 19, 2017 at 21:42
  • $\begingroup$ That's completely true; C is supposed to depend on the derivative bound. I edited to try to make it clearer. Can anyone suggest a better title? $\endgroup$
    – Jason Cantarella
    Oct 19, 2017 at 21:43
  • $\begingroup$ The largest possible quotient $\|f\|_{\infty}/\|f\|_1$ for the ball (and for a given $K=\|\nabla f\|_{\infty}$) is obviously achieved by $f(x)=\|f\|_{\infty}-K|x-a|$, with $a$ from the boundary. That gives you the optimal $C(K,B)$ in principle, though I'm not sure it'll be possible to find $\|f\|_1$ explicitly. $\endgroup$
    – Christian Remling
    Oct 19, 2017 at 23:25

1 Answer 1

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Suppose $D$ is star-shaped and bounded. Fix $x \in D$ such that the line segment joining $x$ to any other point in $D$ lies entirely in $D$. Let $r = \sup \{|x - y|: y \in D\}$ and $d = \sup \{|\nabla f(y)|: y \in D\}$.

If $c = f(x)$ then everywhere on $D$ the values of $f$ lie between $c - rd$ and $c + rd$. So if $\|f\|_1 = 1$ then $(c - rd)Vol(D) \leq 1$ and $(c + rd)Vol(D) \geq -1$. This entails the bound $\|f\|_\infty \leq \frac{1}{Vol(D)} + 2rd$. By scaling this yields the inequality you want with $C = \frac{1}{Vol(D)} + 2rd$.

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  • $\begingroup$ So, this is a cool argument, but I don't see the "by scaling" part. I agree that $\endgroup$
    – Jason Cantarella
    Oct 23, 2017 at 19:24
  • $\begingroup$ Given arbitrary $f \in L^1(D)$, define $g = \frac{f}{\|f\|_1}$. Then the argument I gave shows that $\|g\|_\infty \leq \frac{1}{Vol(D)} + 2rd$, and putting $\frac{f}{\|f\|_1}$ for $g$ in this yields $\|f\|_\infty \leq (\frac{1}{Vol(D)} + 2rd)\|f\|_1$. $\endgroup$
    – Nik Weaver
    Oct 23, 2017 at 20:07
  • $\begingroup$ Proposed counterexample: $f(x) = x^2/5$ on the interval $[-1,1]$. $||f||_\infty = 1/5$, $d = 2/5$, $r = 1$, $\operatorname{Vol}(D) = 2$. I get $\frac{1}{\operatorname{Vol}(D)} + 2 r d = 13/10$ and $||f||_1 = \int_{-1}^1 x^2/5 \, dx = 2/15$. But then the right hand side is $13/10 \times 2/15 = 13/75$, which is surely less than $||f||_\infty = 1/5$. $\endgroup$
    – Jason Cantarella
    Oct 24, 2017 at 14:02
  • $\begingroup$ I think your argument actually proves $||f||_\infty <= \frac{||f||_1}{\operatorname{Vol}(D)} + 2 r d$, which would certainly be true for the function above. $\endgroup$
    – Jason Cantarella
    Oct 24, 2017 at 14:07
  • $\begingroup$ Oh, you're right, because the value of $d$ changes when we scale $f$. $\endgroup$
    – Nik Weaver
    Oct 24, 2017 at 14:52

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