## A Cute Problem in Undergrad Analysis

*Date Written: July 14, 2023; Last Modified: July 14, 2023*

This problem was shared on a Discord server, and according to the guy that posted it, he was given the problem on an undergraduate analysis exam. I’ll share the approaches I tried, as well as my (rather convoluted) solution at the end. An elementary solution still eludes me, please tell me if you come up with one.

**Problem 1. **

Let \(f\) be a function satisfying \(0\leq f(x)\leq f(y)\) if \(x\geq y\). Suppose \(\int_0^\infty f(x) dx\) converges. Prove \(\lim_{x\to\infty} xf(x)=0\).

We should note first that the condition on \(f\) is just saying that \(f\) is nonnegative and nonincreasing.

### Some Initial Attempts

At first glance, it’s tempting to proceed by contradiction. If \(\lim_{x\to\infty} xf(x)=\epsilon>0\), then \(f(x)\) is close to \(\frac\epsilon x\) for large \(x\); hence the integral shouldn’t converge. If \(xf(x)\) was also nonnegative and nonicreasing, this would actually work, as the limit is guaranteed to exist (or approach \(\infty\)), but we don’t know that. This argument thus misses the case where \(\lim_{x\to\infty} xf(x)\) does not exist.

To amend this, one may try to consider the case where \(\limsup_{x\to\infty} xf(x)>0\). The \(\liminf\) is certainly nonnegative, so if the \(\limsup\) is zero, everything falls into place. Unfortunately, this only gives you inequalities of the form \(xf(x)\leq\cdots\), which isn’t very helpful for this problem.

Another approach is to consider the change of variables \(x=e^y\), hence \(dx=e^y dy\). This shows that the integral \(\int_1^\infty e^y f\left(e^y\right) dy\) converges; from here, it’s also tempting to argue that \(\lim_{x\to\infty} xf(x)=\lim_{y\to\infty} e^y f\left(e^y\right)=0\). But again, convergence of the integral does not directly imply this limit; we would need \(e^y f\left(e^y\right)\) to be nonincreasing for that to work.

And finally, the third approach I considered was integrating by parts; this would give you \[ \int_0^\infty f(x)dx=\left.xf(x)\right|_0^\infty-\int_0^\infty x f’(x) dx.\] Perhaps one could then do some funny junk to estimate \(x f(x)\). Unfortunately, \(f(x)\) may not be differentiable (or even continuous), rendering this formula invalid.

### My Solution

Geometrically, the quantity \(t f(t)\) represents the area of the rectangle with corners at \((0, 0)\) and at \((t, f(t))\); this rectangle lies “underneath” the curve. We wish to show that the area of this rectangle approaches \(0\) as \(t\to\infty\).

We may bound the area of the rectangle as follows: define the function \[f_t(x) = \min\left\lbrace f(t), f(x)\right\rbrace.\] Because \(f\) is nonicreasing, \(f_t(x)=f(t)\) for all \(x\leq t\). Crucially, since \(f(x)\) is monotone and bounded from below it has a limit; since its integral converges we necessarily have \(\lim_{x\to\infty} f(x)=0\). Hence, \(\lim_{t\to\infty} f_t(x) = 0\) for all \(x\).

We have that \(f_t(x)\) is dominated by \(f(x)\), which is absolutely integrable, so the dominated convergence theorem says \[\lim_{t\to\infty}\int_0^\infty f_t(x) dx = \int_0^\infty \lim_{t\to\infty} f_t(x) dx = 0.\] (To be more precise, we should pick a sequence of \(t\)’s and use the monotonicity of that integral to justify this limit.) Draw a picture of what this signifies: it will become clear that \[0\leq tf(t)\leq \int_0^\infty f_t(x) dx\] for all \(t\), and the squeeze theorem lets us conclude that \(\lim_{t\to\infty} tf(t)=0\).