# A uniformly continuous function whose integral $\int_0^\infty f(x)dx$ exists converges to zero

Let $f$ be uniformly continuous on $[0,\infty)$ and assume that $\int_0^\infty f(x)dx$ exists. Prove that $$\lim_{x\rightarrow\infty}f(x)=0.$$

This seems obvious, but I couldn't prove it.

Edit Thanks to the hint by Did, I managed to prove the theorem (See the answer below).

• $\int_0^\infty\sin(x^2)\,dx$ exists, but the function is not uniformly continuous. Thus, the limit as $x\to\infty$ is nonzero. In other words, you will need to use the fact that the function is uniformly continuous. Sep 21, 2015 at 0:10
• "How can I show the limit must exist?" This would be only a partial result. Why not work by contradiction, assuming that $|f(x_n)|\ge\epsilon$ for some sequence $x_n\to\infty$. Then use uniform continuity around each $x_n$, hence...
– Did
Sep 21, 2015 at 0:21
• You may apply Barbalat's lemma to the function $F(t)=\int_0^t f(x)\,dx$.
– A.Γ.
Sep 21, 2015 at 0:26
• @Did okay, I got it. Thanks! Sep 21, 2015 at 0:30

Suppose $$f$$ does not converge to $$0$$. Then for a $$\epsilon>0$$, we can always find a sequence $$x_n\to\infty$$ such that $$|f(x_n)|>\epsilon\tag{1}$$ Since $$f$$ is uniformly continuous, there is a $$0<\delta<1$$ that $$|f(x)-f(y)|<\epsilon/2,\quad\text{ whenever }\quad |x-y|<\delta$$

Let $$I_n=[x_n-\delta/2,x_n+\delta/2]$$. So for any $$x\in I_n$$, by $$(1)$$ there is $$f(x)>f(x_n)-\epsilon/2>\epsilon/2\tag2$$ And by $$(2)$$, we have $$\biggl|\,\int_{I_n} f(x)\, dx\,\biggr|\geqslant \frac{\epsilon}{2}\cdot \delta$$ for each $$n$$. But by the Cauchy criterion for integral, $$x_n\to\infty$$ implies that $$\int_0^\infty f(x)\,dx$$ diverges, contradiction. Thus we must have $$\lim\limits_{x\to\infty}f(x)= 0$$.

• Your first line should say that the function does not converge to zero (it might not converge at all).
– Ian
Sep 21, 2015 at 1:23
• Oh, it took me so long to write my proof Latex-wise, when I posted it I saw yours. Sep 21, 2015 at 1:26
• It looks good. Thanks a lot! Sep 21, 2015 at 1:26

Thanks to Did, I managed to prove the theorem.

By contradiction, let's say $f$ doesn't converge to $0$. So by Heine's characterization of a limit, there exists $\epsilon_0$ and a sequence $x_n\rightarrow\infty$ such that $|f(x_n)|\geq \epsilon_0$.

By uniform convergence, for $\epsilon_0$ and for every $x_n$ in the sequence there's a corresponding $\delta_n$ such that for every $x\in(x_n-\delta_n,x_n+\delta_n)$ we have $|f(x_n)-f(x)|<\epsilon_0$.

By the Cauchy criterea, the integral $\int_0^\infty f(x)$ exists iff for every $\epsilon>0$ there is an $A>0$ such that for every $a,b>A$ we have $|\int_a^b f(x)|<\epsilon$.

So we will apply this for $\epsilon=2\epsilon_0 \sup\{\delta_n\}$. Let $x_0$ be the corresponding $A$.

Now, we'll take an $x_n$ in our sequence such that $x_n-\delta_n>x_0$.

We get the following:

$\int_{x_n-\delta_n}^{x_n+\delta_n} f(x)dx \geq \int_{x_n-\delta_n}^{x_n+\delta_n}\epsilon_0dx=2\delta_n\epsilon_0\geq\epsilon$

But this contradicts the existence of the integral by the Cauchy criterea.

Does that look alright?

• The $\delta$ here does not vary with $n$. Suffice to say that without the uniform continuity, and only continuity, the $\delta$ in this proof can vary with $n$. Sep 1, 2016 at 22:05