If $g$ is uniformly continuous and $f(x)$ is close to $g(x)$ (for large enough $x$), is then $f$ uniformly continuous?

Question

Suppose $f$ is continuous on $(0,\infty)$ and $f$ is simmilar $g$ for all $x>M$ ($M>0$).

(i.e. for any $\epsilon >0$, there is $M>0$ such that if $x>M$, then $|f-g|< \epsilon$)

is it true that

if $g$ is uniformly continuous on $(0,\infty)$, the $f$ is also uniformly continuous on $(0,\infty)$.

if $g$ is not uniformly continuous on $(0,\infty)$, the $f$ is not uniformly continuous on $(0,\infty)$.

motivate this question is

"$f(x)=x^3 \sin(\frac{1}{x})$ is uniformly continuous?"

($f(x)$ is similar $x^2$ as $x \rightarrow \infty$ and $x^2$ is not uniformly continuous, I think $f$ is not uniformly continuous.)

Answer 1

I assume you mean "uniformly continuous on $(0,\infty)$, since otherwise, your variable $M$ is overloaded.

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For your particular function, the easiest way to show it is not uniformly continuous is to show that $f'(x)\to\infty$ as $x\to\infty$. Such a function cannot be uniformly continuous because of the mean value theorem.

That said, for your general question, what you are asking is not true. A counterexample is $f(x)=\frac1x$ and $g(x)=0$.

Answer 2

For proving $f(x)= x^3\sin\frac1x$ is not uniformly continuous on $(0,+\infty)$ using the definition, note that if $x-y>\delta/6>0$ then we have \begin{align}f(x)-f(y)&>x^3\left(\frac1x-\frac{1}{6x^3}\right)-y^2 \\ &=x^2-y^2-1/6> \frac16 (\delta(x+y)-1).\end{align}

By the way, even if we assume $f(x),g(x)\to+\infty$ your general claim doesn't hold. Consider $f(x)=\frac1x+x^2 \sin\frac1x, g(x)=x$.

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However,if $f,g:\mathbb{R^m}\to\mathbb{R^n}$ are globally continuous and satisfy $\lim\limits_{\underline{x}\to\infty}f(\underline{x})-g(\underline{x})=0$, then yes, the uniform continuity of one is equivalent to that of the other.