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Let $f :(0, 1) \to \mathbb{R}$ be continuous. Pick out the statements which imply that $f$ is uniformly continuous.

a. $|f (x) − f (y)| ≤ \sqrt{|x − y|}$, for all $x, y ∈ (0, 1) $.

b. $f\left(\frac{1}{n}\right) \to \frac{1}{2}$ and $f\left(\frac{1}{n^2}\right) \to \frac{1}{4}$.

c. $f(x) = x ^{\frac{1}{2}}\sin\left(\frac{1}{x^3}\right)$.

My thoughts:
(c) is uniformly continuous as limits at $x=0$ and $x=1$ are exists.
(a) and (b) I am not sure.

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  • $\begingroup$ Do you know the difference between continuous and uniformly continuous? $\endgroup$
    – Erick Wong
    Jan 5, 2013 at 7:27

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(a) Do you really know the definition of uniform continuity? This part is so obvious.

(b) Consider $\left\vert f\left(\frac1{n^2-1}\right)-f\left(\frac1{n^2}\right)\right\vert$. Note that $\left\vert \frac1{n^2-1}-\frac1{n^2}\right\vert$ will be smaller than any prespecified $\delta>0$ when $n$ is large enough.

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  • $\begingroup$ Given that $f(1/n^2)$ is a subsequence of $f(1/n)$, I'm tempted to say the contradiction in (b) vacuously implies that $f$ is uniformly continuous. $\endgroup$
    – Erick Wong
    Jan 5, 2013 at 18:12
  • $\begingroup$ @ErickWong Absolutely agree! Yet I think part (b) was intened to mean that $f(1/n)\to1/2$ for the sequence of nonperfect squares. $\endgroup$
    – user1551
    Jan 5, 2013 at 18:43

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