Domain: $(0, \infty)$

I have $2$ functions:

$$ f(x) = \sqrt{x}, \quad g(x) = x \cdot \sin(1/x) $$

The answers say that $f(x)$ is uniformly continuous because at $0$ it has a finite limit and in the $\infty$ its derivative is bounded.

For $g(x)$, it has finite limits at the boundaries of the interval, namely at $0$ and $\infty$ and therefore it is uniformly continuous.

Can someone explain to me how those facts prove that the functions are uniformly continuous?

I am familiar with the formal definition and with the fact that if the derivative is bounded in the interval than the function is uniformly continuous (for continuous functions).

  • 2
    $\begingroup$ You should clearly specify the domain of these functions. $\endgroup$ – zhw. May 23 at 17:31

Consider a continuous function $f:(a,b) \rightarrow \mathbb{R} $ with $-\infty \le a < b \le \infty$. What the answers uses is that when $\lim\limits_{x \rightarrow a} f(x)$ and $\lim\limits_{x \rightarrow b} f(x)$ exists, then $f$ is uniform continuous. To understand this, note that by existence of $f := \lim\limits_{x \rightarrow a} f(x)$ we can find for all $\varepsilon >0$ some $b'<b$ such that for all $x\ge b'$ we have $|f(x) - f| < \varepsilon$. Therefore $|f(x) - f(y)| < 2\varepsilon$ for all $x,y\ge b'$. Analogously we can find some $a' > a$. Now $f$ is uniform continuous on the compact set $[a',b']$ and you combine this with previous observation to get uniform continuouty on (a,b). Take a look here.

Now in you first example, as you said, $|f'(x)|$ is bounded on $[1, \infty)$ and therefore $f$ is uniform continuous on $[1, \infty)$. Moreover, by the above reasoning $f(x)$ is also uniform continuous on (0,1). Here, we can also use that $[0,1]$ is compact and hence $f$ is uniform contionuous.

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