Prove f(x) times a uniformly continuous fucntion is not uniformly continuous

Question

I know the proof for $x$sinx is not uniformly continuous. In that proof, it uses a property of sinx that sin$(2\pi n)=0$. Then what if there is a uniformly continuous function $h(x)$. $h(x)$ is periodic and bounded, but $h(x)$ never equals to zero. Then let $f(x)=x$. Prove $h(x)f(x)$ is not uniformly continuous.

$h(x)$ is a uniformly continuous and periodic function that is greater than zero. prove $xh(x)$ is not uniformly continuous

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Edit：

$h(x)$ is a uniformly continuous and periodic function that is greater than zero and h(x) is not a constant funtion. prove $xh(x)$ is not uniformly continuous

This is my attempt:

Since $h(x)$ is bounded and continuous, it takes on its minimum. Suppose $h(a)=\beta,a>0$ is the minimum of $h(x)$

Then the thing that needs to prove is

$\exists \epsilon,\forall\delta>0,\exists x,x_0\in$ domain such that $|x-x_0|<\delta$ and$|xh(x)-x_0h(x_0)|\geq\epsilon$

Then let $0<x_0<x$ take $x_0=a$ and $x=a+\frac{1}{n}$,so $|x-x_0|=x-x_0=\frac{1}{n}<\delta$

$|xh(x)-x_0h(x_0)|=|(a+\frac{1}{n})h(a+\frac{1}{n})-ah(a)|=|(a+\frac{1}{n})h(a+\frac{1}{n})-a\beta|$

Since $h(a+\frac{1}{n})\geq h(a)=\beta$

Then $(a+\frac{1}{n})h(a+\frac{1}{n})\geq (a+\frac{1}{n})\beta> a\beta$

Thus $|(a+\frac{1}{n})h(a+\frac{1}{n})-a\beta|=(a+\frac{1}{n})h(a+\frac{1}{n})-a\beta>0$

Then there should exist a desired $\epsilon$ to satisfy this.

However, I don't believe my proof is correct, and since there's no other quantifier before $\epsilon$, I think the value of $\epsilon$ should not depend on anything. Thus the value of $\epsilon$ should be a specific value. As a result, I think I'm stuck. Any helps? Thanks in advance.

Answer 1

Since $h$ not constant, for arbitrarily small $\delta$ there exist $q>0$ and (by periodicity) arbitrarily large $x$ such that $h(x+\delta) - h(x) = q$. Then, since $h$ is positive, $$(x+\delta)h(x+\delta) - xh(x) \geq xh(x+\delta) - xh(x) = qx.$$ As $x$ can be taken arbitrarily large the expression on the right-hand side can be made e.g. greater than 1, thus contradicting uniform continuity.

The problem with your reasoning is that you take $h(a)\leq h(a+\frac 1n)$ instead of the strict inequality, thus not using the fact that $h$ is non-constant.