Alternative way to show $f(x) = x^2$ is uniformly continuous in a fixed closed interval

Question

We wish to show $f(x) = x^2$ is uniformly continuous in $[a,b]$. This is the way I did it, but it's different from how it's "supposed" to be done. Thus I naturally have some doubts and so here I am.

Firstly, I want to show that it is continuous in every point $x_0$ in its domain (in general, for the set $\mathbb{R}$). This is so I have some $\delta$ to work with. So for $|x - x_0| < \delta$

$|f(x) - f(x_0)| = |x - x_0||(x-x_0) + 2x_0| ≤ |x - x_0|(|x-x_0| + 2|x_0|) < \delta(\delta + 2|x_0|) = \epsilon$

which is clear if we choose $\delta = \sqrt{\epsilon + |x_{0}|^2} - |x_{0}|$

Now, we show it is uniformly continuous. Fix some $\epsilon > 0$. Note that $f(x) = \sqrt{\epsilon + x^2} - x$ is monotonically decreasing for $x>0$ and fixed $\epsilon > 0$ (this is easily shown by differentiating). Either we have $|a| > |b|$ or $|a| ≤ |b|$. In either case, it is fairly clear that for all $x_0 \in [a,b]$, $|x_0| < \max(|a|, |b|)$.

If it is the first case, we can set $\delta = \sqrt{\epsilon + |a|^2} - |a|$ as the uniform modulus of continuity since it is sufficiently small (particularly, it is smaller than $\sqrt{\epsilon + |x_{0}|^2} - |x_{0}|$ for $x_0 \in (a,b]$) and in the second, similarly, we can set $\delta = \sqrt{\epsilon + |b|^2} - |b|$.

Answer 1

The continuity of $x^2$ is used to show that the square root function exists (via the intermediate value theorem). So unless you are accepting that the function is at least continuous to begin with, you are making a circular argument by using square roots. In other respects the proof is correct.

To prove that $g(x) = \sqrt{\epsilon + x^2} - x$ is decreasing, you can also write $g(x) = \epsilon\big/\left(\sqrt{\epsilon + x^2} + x\right)$ instead of differentiating.

Since you are approaching this from an advanced viewpoint (using differentiation), you might be interested in trying to prove the following lemma: if $f'(x)$ is bounded on an interval, then $f(x)$ is uniformly continuous there. To prove this, first show that $|f'(x)| \leq M$ implies $|f(s) - f(t)| \leq M(s-t)$.

Answer 2

We can show directly uniform continuity of $f$ using the fact that it attains a maximum over a closed interval. Let $M = \text{max}(|f(x)|: x \in [a,b]) > 0$, and let $\epsilon > 0$ be any positive real number, then choose $\delta = \dfrac{\epsilon}{2M} > 0$, thus for any $x,y \in [a,b]$ and if $|x-y| < \delta \Rightarrow |f(x) - f(y)| = |x^2-y^2| = |x+y||x-y| \leq (|x|+|y|)|x-y| \leq 2M|x-y|< 2M\cdot \delta = \epsilon$. This shows that $f$ is uniformly continuous on $[a,b]$.