Let $f: (0, 1]\to \mathbb{R}$ be continuous on the domain. I want the condition of $f(x)$ where $g(x) = f(x)\sin(1/x)$ is uniformly continuous on $(0, 1]$. I expect that the answer would be $\lim_{x\to\ 0^{+}}f(x)=0$.
When $\lim_{x\to\ 0^{+}}f(x)=0$, and define $f(0)$ as $0$, $f(x)$ becomes continuous by the definition, and since $f(x)$ is continuous on [0, 1] which is compact, $f(x)$ is uniformly continuous at (0, 1].
However, I want to know whether the converse holds. That is, if $g(x)$ is uniformly continuous, is $\lim_{x\to\ 0^{+}}f(x)=0$ ???
Thank you for your help!!
We have the well known theorem: A function $g$ continuous on $(a,b)$ is uniformly continuous if and only if $$\lim_{x\to a^+} g(x) \quad \text{and} \quad \lim_{x\to b^-} g(x)$$ exists.
Because $g(x)$ is given as uniformly continuous on $(0,1]$ we can deduce $$\lim_{x\to 0^+}g(x)$$ exists. Additionally we get that the unique limit has to be $0$ by considering the sequence given by $$x_n = \frac{1}{2\pi n}$$
Now it's easy to see that if $\lim\limits_{x\to\ 0^{+}}f(x)$ exists it has to hold $$\lim_{x\to\ 0^{+}}f(x)=0$$ but only IF the limit exists what is for example the case if $f$ is continuous itself in $x=0$.
Unfortunately this does not necessarily hold if the limit does not exists. To see this consider that $$\sin\left(\frac{1}{x}\right)$$ oscillates between $-1$ and $1$ for $x\to 0$ so you will find intervals $I_1 > I_2 > \ldots > 0$ s.t. $$\sin\left(\frac{1}{x}\right) \le 2^{-n} \quad \text{ for } x \in I_n$$
Where I note $I_1 > I_2$ if for all $x \in I_1, y \in I_2$ it holds $x > y$
Now take $f \equiv 0$ for $$x \not\in \bigcup_{n\in\Bbb N} I_n$$ and let f peak continuously to $\frac{1}{4}$ within each $I_n$
Then $g(x) \to 0$ but $\lim_{x\to 0} f(x)$ does not exist by construction.