Let $f: \Bbb R \to [0,\infty)$ be a continuous function such that $g(x)=(f(x))^2$ is uniformly continuous. Which of the following is always true?

Solution 1:

Hint. You want to prove that $f(x) = \sqrt{g(x)}$ is uniformly continuous. The statement of this looks like "$|f(x) - f(y)| \leq \epsilon$ whenever...(condition on $x$ and $y$ being close enough)." Since $g$ is known to be uniformly continuous, you can always write "$|g(x) - g(y)| \leq \alpha$ whenever..."

So it will be enough for you to find some function $h \colon (0,+\infty) \to (0,+\infty)$ with the property that whenever $|g(x) - g(y)| \leq \alpha = h(\epsilon)$, you necessarily have $|f(x) - f(y)|\leq \epsilon$. Since the only thing you really know is that $f(x)$ is the square root of $g(x)$, this clarifies what conditions you would want the function $h$ to satisfy. Now your job is to find a function $h$ that works.

Essentially, the proof amounts to showing that the square root function is uniformly continuous.

Edit. Do you have a theorem that says a composite of uniformly continuous functions is uniformly continuous? If so, all you need to do is prove that $j(t) = \sqrt{t}$ is uniformly continuous, since $f(x) = j(g(x))$. (If not, you will need to include a small additional step that amounts to proving this is the special case you're dealing with.)

To prove the uniform continuity of $j$, it will be enough to prove that, for $a, b \geq 0$, $$|a - b| \leq \epsilon^2 \Longrightarrow |\sqrt{a} - \sqrt{b}| \leq \epsilon.$$

An alternative is to show that $j$ has a Lipschitz constant on $[1,+\infty)$ and to note that $[0,1]$ is a closed bounded interval, hence $j$ must be uniformly continuous there.