Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.

Now, I would like to ask the following question:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.

How do we go about doing it?

share|improve this question

4 Answers 4

$g$ is clearly continuous, you really only need to worry about what would happen at the endpoints. some sequence $x_n\to 1$ has $g(x_n)\to g(1)$, similarly for zero. so you can define $g$ on $[0,1]$ where it will be uniformly continuous (basically $g$ doesnt zoom off to infinity at the endpoints, else $g(x_n)$ wouldnt be cauchy)

share|improve this answer

Just a remark: a key point is that the open interval $(0,1)$ is not complete as a metric space. Suppose you have a function $f: X \rightarrow \mathbb{R}$ where $X$ is complete. Then $f$ is Cauchy-continuous -- i.e., if $\{x_n\}$ is a Cauchy sequence in $X$, then $\{f(x_n)\}$ is Cauchy in $\mathbb{R}$ -- iff it is continuous. To see that this is in general much weaker than being uniformly continuous, consider the example $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto x^2$.

In the other direction, if $X$ is incomplete, but with compact completion, then every Cauchy continuous function $f: X \rightarrow \mathbb{R}$ extends to a continuous function $\overline{f}: \overline{X} \rightarrow \mathbb{R}$ on the completion $\overline{X}$. Since $\overline{X}$ is compact, $\overline{f}$ is uniformly continuous, hence so is its restriction $f$. This latter situation obtains in the question asked by the OP.

share|improve this answer

You have to prove is that if $(x_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ and $(y_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ converge to the same limit $x \in [0,1]$, then $(g(x_n))_{n \ge 0}$ and $(g(y_n))_{n \ge 0}$ have the same limit (you may define $z_{2n} = x_n$, $z_{2n+1} = y_n$, and use uniqueness of the limit). Then you can the deduce that $g$ is continuous on the compact $[0,1]$ (or can be extended to such a function), hence uniformly continuous.

share|improve this answer

You can also prove it by contradiction.

Suppose that $f$ is not uniformly continuous. Then there exists an $\epsilon >0$ so that for each $\delta>0$ there exists $x,y \in (0,1)$ with $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$.

For each $n$ pick $x_n, y_n$ so that $|x_n-y_n| < \frac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \epsilon$.

Pick $x_{k_n}$ a Cauchy subsequence of $x_n$ and $y_{l_n}$ a Cauchy subsequence of $y_{k_n}$.

Then the alternaticng sequence $x_{l_1}, y_{l_1}, x_{l_2}, y_{l_2},..., x_{l_n}, y_{ln}, ...$ is Cauchy but

$$\left| f(x_{l_n}) - f(y_{l_n}) \right| \geq \epsilon \,.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.