Verify: $x \mapsto \sqrt x$ is uniformly continuous on $[0, \infty]$

Question

Prove $x \mapsto \sqrt x$ is uniformly continuous on $[0, \infty]$.

Note: There are more elegant proofs readily available, such as here. This question asks to verify and critique my proof.

Proof: Let $\delta = \varepsilon^2$. Then, for $x, a \geq 0; |x-a| < \delta$, we have:

$$\begin{align*} | \sqrt x - \sqrt a| &= \frac {|x-a|} {\sqrt x + \sqrt a} \\ &\leq \frac {|x-a|} {\sqrt{|x-a|}} && \text{(since $\sqrt x + \sqrt a = \sqrt x + \sqrt{x+a-x} \geq \sqrt{|x-a|}$)} \\ &\leq \sqrt{|x-a|} \\ &< \sqrt \delta \\ & < \varepsilon. \end{align*}$$

Discussion: The other proofs I've seen use one of two approaches:

1. Split the domain into $[0,k]$ and $[k, \infty)$; the first is compact and the second easier to prove, either by bounding the derivative or by the inequality $|\sqrt x - \sqrt a| < |x-a|/2\sqrt k$.
2. Square the left side and apply the triangle inequality for positive quantities to get $|\sqrt x - \sqrt a|^2 \leq |\sqrt x - \sqrt a||\sqrt x + \sqrt a| = |x-a|.$

I imagine those techniques are more broadly applicable. How would I have thought of either? Especially #2, which seems to come from nowhere, the only way I can see myself using it is by seeing it before and recalling the technique.

Answer 1

Generally, proving continuity does require a certain amount of practice and some foresight, as there isn’t really a standardized approach to it.

Regarding the 2nd approach you mentioned: When attempting to prove that a certain function is (uniformly) continuous, you would want to transform the expression $|f(x) - f(x_0)|$ such that it is written as a combination containing the term $|x - x_0|$. That is, once you have reached that point, it is rather straightforward to choose an appropriate $\delta$ in case that is possible, i.e. the function is effectively continuous on the given domain. Applied to the example given by you, the proof for continuity could look like the following:

We start by estimating the absolute value of the difference in function values for all $0 \leq x < \infty$ and $x_0 > 0$ as given in the definition of continuity:

$|f(x) - f(x_0)| = |\sqrt{x} - \sqrt{x_0}| \\ = |\sqrt{x} - \sqrt{x_0}\cdot \frac{\sqrt{x} + \sqrt{x_0}}{\sqrt{x} + \sqrt{x_0}}| \\ = |\frac{x - x_0}{\sqrt{x} + \sqrt{x_0}}| \leq \frac{|x - x_0|}{\sqrt{x_0}} $

We now have attained the above described situation such that for any $\epsilon > 0$, we may choose $\delta = \sqrt{x_0} \cdot \epsilon$, because then we have:

$|x - x_0| < \delta \Rightarrow |f(x) - f(x_0)|< \frac{\delta}{\sqrt{x_0}} = \frac{\sqrt{x_0}}{\sqrt{x_0}}\cdot \epsilon = \epsilon$

In order to show continuity at $x_0 = 0$ as well, we note:

$|f(x) - f(0)| = \sqrt{x} < \epsilon$

We have now showed that $\sqrt{x}$ is continuous on $[0,\infty)$.

If we further want to show uniform continuity, it is easy to see that the implication works out for $\delta = \epsilon^2$, because then we have $|x - x_0| < \delta$ and since we would like to end up having $|\sqrt{x} - \sqrt{x_0}| < \epsilon$, we now square the left side and evaluate:

$|\sqrt{x} - \sqrt{x_0}|^2 \leq |\sqrt{x} - \sqrt{x_0}||\sqrt{x} + \sqrt{x_0}| \\ = |x - x_0| < \delta = \epsilon^2 \Rightarrow \ |\sqrt{x} - \sqrt{x_0}| < \epsilon$

Hence, we conclude that $\sqrt{x}$ is uniformly continuous on $[0,\infty)$.