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.

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  • $\begingroup$ I got a clue from your answer! I already know that if g(x) is uniformly continuous, the image of Cauchy sequence by g is Cauchy sequence. So, when I consider the Cauchy sequence given by x_n = 2/(npi), f(2/(npi)) * sin(npi/2) should be Cauchy sequence, which leads to the fact that lim(x->0+) f(x) = 0, since sin(npi/2) has the value 0, 1, -1 infinitely. Do you think it is correct? Thank you:) $\endgroup$ – Jihun Kim May 3 at 15:27
  • $\begingroup$ I don't think the limit exists necessarily. I will add a counterexample to my answer. $\endgroup$ – Gono May 3 at 15:29
  • $\begingroup$ @mathcounterexamples.net The $f$ sketched is continuous on $(0,1]$ as well as $\sin(\frac{1}{x})$, so $g(x) = f(x)\sin(\frac{1}{x})$ is too. Additionally $$\lim_{x\to 0} g(x) = 0$$ by construction so we have: $g$ continuous on $(0,1]$ and $\lim_{x\to 0} g(x)$ exists, hence $g$ is uniformly continuous on $[0,1]$ (and so especially on $(0,1]$) as mentioned in the beginning of my post. $\endgroup$ – Gono May 3 at 16:52
  • $\begingroup$ $f$ is continuous on $(0,1]$ but not on $[0,1]$… but I never said f is continuous in $x=0$. Actually this was the goal to construct a function $f$ continuous on $(0,1]$ where $$\lim_{x\to 0} f(x)$$ does not exist… so $f$ cannot be continuous on $[0,1]$ otherwise it follows (as shown in my answer) that $$\lim_{x\to 0} f(x) = 0$$ But although $f$ is not continiuous in $x=0$ the limit for $g$ in 0 still exists! $\endgroup$ – Gono May 3 at 17:03
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    $\begingroup$ @Gono You’re right. I’ll delete my previous comment. $\endgroup$ – mathcounterexamples.net May 3 at 17:13

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