If $f(x)$ is continuously decreasing and $\lim_{x \to \infty} f(x) = 0$, is $xf(x)$ uniformly continuous?

Question

Suppose $f:[0, \infty) \to [0, \infty)$ is decreasing and continuous with $\lim_{x \to \infty} f(x) = 0$. Let $g(x) = xf(x)$. Is $g(x)$ uniformly continuous on $[0, \infty)$?

My work: I've been able to come up with a counterexample where $g(x)$ is not Lipschitz. Since Lipschitz is stronger than uniform continuity, this isn't a full solution, but nevertheless the ideas may be useful. That said, there might be a very simple counterexample that's eluding me.

First, recall that there are continuous functions $h:[0, \infty) \to [0, \infty)$ such that $$\limsup_{x \to +\infty} \ h(x) = +\infty$$ but $\int_0^{+\infty} h(x) \ dx < +\infty$

Here is a brief construction: for integer $n \geq 1$, in each interval $[n, n+\frac{2^{1-n}}{n}]$ the graph of $h$ looks like an isosceles triangle of area $2^{-n}$ and height $n$. Elsewhere $h(x) = 0$. It's easy to see the integral of $h$ over $\mathbb{R}_+$ is $1$, but its $\lim \sup$ is $+\infty$.

Now consider that $$f(x) = 1 - \int_0^{x} h(t) \ dt$$

is continuous (actually differentiable), decreasing and vanishing at $+\infty$.

If we consider $g(x) = xf(x)$, note that $g'(x) = xf'(x) + f(x) = f(x) - xh(x)$. $f(x)$ is bounded and $|xh(x)|$ can be made arbitrarily large, and hence $|g'|$ can be made arbitrarily large, so there is no uniform bound on $\left|\frac{g(x) - g(y)}{x-y}\right|$. This implies $g$ is not Lipschitz. This example may or may not be uniformly continuous, but I can't prove it.

Answer 1

Choose a strictly increasing sequence of integers $0=a_0<a_1<\cdots$. Now define $f$ as follows. If $a_k\leq x\leq a_{k+1}-1$ for some $k$ then $f(x)=2^{-k}$; on the interval $[a_{k+1}-1,a_{k+1}]$ the function is linear.

Now on the interval $(a_{k+1}-1,a_{k+1})$ we have $g(x)$ is differentiable with derivative at most $2^{-k}-2^{-k-1}(a_{k+1}-1)$; call this bound $b_k$. If we choose $a_k$ increasing sufficiently quickly, $b_k\to-\infty$. Thus for $x_k$ in the middle of this interval and any given $\varepsilon>0$ we need $\delta\leq\varepsilon/|b_k|$, so no fixed $\delta>0$ works.

Answer 2

Let $a_n=\frac{\ln(1)}{n^{1/2}}$ and $b_n=\sum_{k=2}^n (\ln(k))^{-1}$ for $n>3$. Let $f(b_n)=a_n$ and $f$ have a straight line graph between $b_n$ and $b_{n+1}$. Then f is defined on $[c,\infty)$ for some $c>0$ and $f(x+c)$ has the desired properties.