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When I check uniform continuity of a (continuous) function, I generally look for a part where the slope diverges to infinity (e.g. $y = x^2$, $y = \frac{1}{x}$).

For $f(x)= \sqrt{x}$ for $ x\in [0, \infty)$, I found that the slope diverges to infinity as $x$ approaches $0$.

So, I tried to prove $f$ is not uniformly continuous by setting up two observation points near 0.

Define $x_n = \frac{1}{n^2}$ and $y_n = \frac{4}{n^2}$.

For any $\delta > 0$, let $\epsilon = \delta$ we can choose $N \in \mathbb{N}$ such that $ N > \sqrt{\frac{3}{\delta}} $.

Then, $|y_n-x_n| = \frac{3}{N^2} < \delta$,

but $|f(y_n)-f(x_n)| = \frac{1}{N} > \frac{3}{N^2} = \delta = \epsilon$ for $N > 3$.

Therefore, $f(x)$ is not uniformly continuous.

Where did I go wrong? Is it misleading to think of uniform continuity as the slope diverging to infinity?

Math solver
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hnbm
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  • I believe your logic is incorrect. To prove not uniformly continuous, you would need to show there exists an epsilon such that no delta's work, i.e. you must choose an epsilon first. – JLinsta Dec 05 '23 at 16:23

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Your reasoning is not correct because it only proves one delta doesn't work, but it doesn't show no delta will work.

It is easy to argue that sqrt(x) is uniform continuous on $[0,1]$ because it's continuous on the compact set as continuous functions on a compact set are uniform continuous.

As for your concern why functions with unbounded derivative can be uniform continuous, I encourage you to read the link above. Intuitively speaking, for fixed $\epsilon$, when $x$ is close to $0$, we will need a smaller $\delta$, but because the domain is compact, we only need finitely many of those deltas, so there is a minimum delta.

Charlie Chen
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