Trying to solve

$f(x)$ is uniformly continuous in the range of $[0, +\infty)$ and $\int_a^\infty f(x)dx $ converges.

I need to prove that: $$\lim \limits_{x \to \infty} f(x) = 0$$

Would appreciate your help!


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  • 1
    $\begingroup$ If the limit exists and is equal to $L\neq 0$, then show that the integral diverges (for example, if $L\gt 0$, then eventually we will have $f(x)\gt \frac{L}{2}$, and the integral will be bounded below by a quantity that diverges to $\infty$). Conclude that if the limit exists, then it must be equal to $0$. Then use uniform continuity to show the limit does in fact exist. $\endgroup$ – Arturo Magidin Dec 16 '11 at 20:35
  • $\begingroup$ the part of using the uniform continuity to show the limit does in fact exist is the part i'm having trouble with :/. $\endgroup$ – Guy Dec 16 '11 at 20:40

If the integral converges than you have: $\lim_{M\to\infty} \int_a^M f(x) dx = c\ $ for some $c$. Assuming that f does not converge to 0 we have: $\forall M>0\ \exists p>M: |f(p)|>\varepsilon\ \ $ for some $\varepsilon$.

Without loss of generality we have $\forall M>0\ \exists p>M: f(p)>\varepsilon\ \ $. So there exists $(p_n)_{n=1}^\infty$ s.t. $p_n \to \infty\ $.

Using uniform continuity we have that for $q: |p-q|<\delta\ \ $ the inequality holds: $f(q)>\varepsilon/2\ $. So $\int_a^{p+\delta} f(x) dx - \int_a^{p-\delta} f(x) dx > \delta \cdot \epsilon \ \ $, which means that the sequence $(\int_a^{q_n}f(x)dx)_{n=1}^\infty$ (where $q_0=p_0-\delta,\ q_1=p_0+\delta,\ q_2=p_1-\delta \ldots)\ \ $ is not a Cauchy sequence so it can not converge.

  • $\begingroup$ How can I prove that the limit of f(x) does in fact exist? I understand how to prove that if it does then it is equal to 0 $\endgroup$ – Guy Dec 16 '11 at 20:48
  • $\begingroup$ @Guy: This is a proof of the exact statement you want by contradiction. $\endgroup$ – Jonas Meyer Dec 16 '11 at 20:50

Suppose $$\tag{1}\lim\limits_{x\rightarrow\infty}f(x)\ne 0.$$

Then we may, and do, select an $\alpha>0$ and a sequence $\{x_n\}$ so that for any $n$, $$\tag{2}x_n\ge x_{n-1}+1$$ and $$\tag{3}|f(x_n)|>\alpha.$$

Now, since $f$ is uniformly continuous, there is a $1>\delta>0$ so that $$\tag{4}|f(x)-f(y)|<\alpha/2,\quad\text{ whenever }\quad |x-y|<\delta.$$

Consider the contribution to the integral of the intervals $I_n=[x_n-\delta/2,x_n+\delta/2]$: We have, by (3), and (4) that $$\biggl|\,\int_{I_n} f(x)\, dx\,\biggr|\ge {\alpha\over2}\cdot \delta$$ for each positive integer $n$.

But, by (2), the $x_n$ tend to infinity. This implies that $\int_a^\infty f(x)\,dx$ diverges, a contradiction.

Having obtained a contradiction, our initial assumption, (1), must be incorrect. Thus, we must have $\lim\limits_{x\rightarrow\infty}f(x)= 0 $.

Take this with a grain of salt, but, informally, the idea used above is based on the following:

For clarity, assume $f>0$, here.

If the integral $\int_a^\infty f(x)\,dx$ is convergent, then for large $x$, the graph of $f$ is close to the $x$-axis "most of the time" and, in fact, the positive $x$-axis is an asymptote of "most of" the graph of $f$.

I say "most of the time" and "most of" because is not necessarily so that a function $f$ which is merely continuous must tend to 0 when $\int_a^\infty f(x)\,dx$ converges. There may be spikes in the graph of $f$ as you go out in the positive $x$ direction. Though the height of the spikes can be large, the width of the spikes would be small enough so that the integral converges (so, the sum of the areas of the spikes is finite).

But the graph of a uniformly continuous function that is "mostly asymptotic to the $x$-axis" does not have very tall spikes of very short widths arbitrarily far out in the $x$-axis.

  • $\begingroup$ Thanks but I still need to prove that the limit does in fact exists $\endgroup$ – Guy Dec 16 '11 at 20:44
  • $\begingroup$ @Guy: To show $P\implies Q$, proof by contradiction uses the approach of showing the equivalent statement that $P\land\neg Q$ leads to a contradiction. Actually this is easily formulated as a proof by contraposition (no need for contradiction), $\neg Q\implies \neg P$. (If you're not familiar with contraposition, Google it.) $\endgroup$ – Jonas Meyer Dec 16 '11 at 20:48
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    $\begingroup$ Note that $\lim_{x \to \infty} f(x) \ne 0$ does not mean "the limit exists and is not 0", it means "it is not true that the limit is 0". $\endgroup$ – Robert Israel Dec 16 '11 at 20:59
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    $\begingroup$ One must note that the assumption $f>0$ is very important to say that $f$ is "most of the time" near to x-axis. Otherwise you can take a function $\sqrt{x}\sin{x^2}$, which is integrable (ask wolfram), but it is not near x-axis "most of the time". $\endgroup$ – savick01 Dec 16 '11 at 22:09

For all $\delta>0$, the fact that $\int_0^\infty f(x)dx=\sum\limits_{k=0}^\infty\int\limits_{k\delta}^{(k+1)\delta}f(x)dx$ converges implies that $\lim\limits_{k\to\infty}\int\limits_{k\delta}^{(k+1)\delta}f(x)dx=0$. Let $\varepsilon>0$ be given. Let $\delta>0$ be such that $|x-y|<2\delta$ implies $|f(x)-f(y)|<\varepsilon/2$. Let $N>0$ be such that $k\geq N$ implies $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|<\frac{\varepsilon\delta}{2}$. Let $x_0\geq (N+1)\delta$. Then there exists $k\geq N$ such that $x_0$ is in $[k\delta,(k+1)\delta]$. It follows that $|f(x)-f(x_0)|<\varepsilon/2$ for all $x\in[k\delta,(k+1)\delta]$, and this implies that $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|\geq \delta(|f(x_0)|-\varepsilon/2)$. Since $\left|\int\limits_{k\delta}^{(k+1)\delta}f(x)dx\right|<\frac{\varepsilon\delta}{2}$, this implies that $|f(x_0)|<\varepsilon$. Therefore $\lim\limits_{x\to\infty}f(x)=0$.


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