Nonnegative function with finite integral and Lipschitz implies bounded?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $f(0)=M$.