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Suppose $f: \mathbb{R} \to \mathbb{R}$ is nonnegative, Lipschitz and has finite integral $\int f \, dx <\infty$. Is $f$ necessarily bounded?

I think it must be: if not, we can find a sequence $x_n$ such that $f(x_n) > n$. And then integrating on $[x_n -1/2, x_n+1/2]$ we see that the integral is at least on the order of $n$, so by taking $n$ large, we get a contradiction.

Is this reasoning accurate, and can we give an explicit bound on $f$ in terms of the Lipschitz constant?

Drew Brady
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3 Answers3

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So $f$ is uniformly continuous and integrable. It suffices to prove that $\lim_{|x|\rightarrow\infty}f(x)=0$, for then $f$ must be bounded, but the answer has been given in here.

user284331
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Your answer is correct. But there cannot be a bound depending only on the Lipschitz constant $C$. For any $M>0$ consider the function $f(x)=(M-|x|)^{+}$. This is Lipschitz with $C=1$ and it is integrable. But $f(0)=M$.

Kavi Rama Murthy
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Presumably you mean globally Lipschitz, suppose with constant $L$.

Suppose $f(x) > 0$ then $\int f \ge {1 \over L} f(x)^2$.

Hence $\sup f \le \sqrt{L \int f}$.

Addendum:

Since $f$ is Lipschitz, we have $f(y)-f(x) \ge -L|x-y|$, or $f(y) \ge f(x) -L|x-y|$.

Now note that $\int_{x-{f(x) \over L}}^{x+{ f(x) \over L}} (f(x) -L|x-y|) dy\ge {1 \over 2} 2 {f(x) \over L} f(x)$.

copper.hat
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    Can you explain the first inequality: $\int f \geq f(x)^2/L$? – Drew Brady Dec 21 '19 at 06:02
  • @DrewBrady: I added an elaboration. It might help to draw a little picture to see the triangle involved. – copper.hat Dec 21 '19 at 06:08
  • @cooper.hat Do you know if this bound is attained? – Drew Brady Dec 22 '19 at 22:37
  • Take $f(x)=\max(0,1-|x|)$, Then $\int f = 1$, $L=1$, $\sup f = 1$. – copper.hat Dec 22 '19 at 23:12
  • Right -- what is the correct modification for $\sqrt{L\int f}$, though? If one tries $f(x) = (1 - L|x|)+$, we obtain $L$ Lipschitz, but bounded by 1. The bound predicts $\sqrt{L}$. If one tries $f(x) = L(1 - |x|)+$, again we have $L$ Lipschitz, but bounded by $L$; the bound predicts $L^{3/2}$, however. – Drew Brady Dec 22 '19 at 23:18
  • Huh? $\sqrt{ 1 \cdot 1} = 1$. I don't follow what you are saying. – copper.hat Dec 22 '19 at 23:32
  • @cooper.hat Yes, I agree that your bound is optimal for $L = 1$. I am asking if your bound is tight in a stronger sense: let $L > 0$. Does there exist $f = f_L$ such that $f$ is $L$-Lipschitz and $\sup f = \sqrt{L \int f}$? – Drew Brady Dec 22 '19 at 23:53