# Let $f:(0,1)→R$ be continuous. Pick out the statements which imply that $f$ is uniformly continuous.

Let $f :(0, 1)→ R$ be continuous. Pick out the statements which imply that $f$ is uniformly continuous.

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

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

• It is even discontinuous in $0$ about the $f$ in $(b)$. And $f$ in $(a)$ satisfies Lipschitz Condition so it is uniformly continuous. Jan 22, 2014 at 8:45
• Domain of function is open (0,1) Jan 22, 2014 at 8:47
• Lemma. $f$ is uniformly in $(a,b)$ $\Longleftrightarrow$ the limits of $a,b$ are exist. Then you can extend the function on whole $[a,b]$. Jan 22, 2014 at 8:51
• If |f(x)−f(y)|≤k.|x−y|, then f satisfies lipscitz condition, in a it is not satisfied- gaoxinge Jan 22, 2014 at 8:52
• (b) can't even happen. If $f(1/n) \to 1/2$, then for $n$ big enough $f(1/n) > 3/8$. In particular for $n$ big enough $f(1/n^2) > 3/8$, and so $f(1/n^2)$ can't converge to $1/4$. Jan 22, 2014 at 8:56

Notice this

$$|f(x) − f(y)| ≤ \sqrt{|x − y|} < \epsilon \implies |x-y|< \epsilon^2 = \delta.$$

• what about option b? Jan 22, 2014 at 8:45
• impossible since $n^2$ is a subsequence of $n$
– Blah
Jan 22, 2014 at 8:48
• Mhenni Benghorbal can you give me a conterexample or proper justification for case b in your answer? Jan 22, 2014 at 9:01

For option $(b)$ does not implies that uniform convergence. Take $a_n=\frac{1}{n}$ and $b_n=\frac{1}{n^2}$.Here $a_n-b_n \rightarrow 0$ but $f(a_n)-f(b_n)\nrightarrow 0$ . Hence (b) does not implies uniform continuity.

• See! suppose (b) implies f is uniformly continuous. Then if we take a seqn $a_n$ and $b_n$ as the above. By uniform continuity $f(a_n)-f(b_n)$ goes to zero. so contradiction. Jan 22, 2014 at 9:06
• It seems to me that "does not imply" is, technically, the wrong answer to part b) in view of nik's comment above. E.g. doesn't $1 = 2$ imply that $f$ is uniformly continuous? Jan 22, 2014 at 18:36