Does the converse of uniform continuity -> Preservance of Cauchy sequences hold?

We recognize that if a function $f$ is evenly continual on an interval $I$ and also $(x_n)$ is a Cauchy series in $I$, after that $f(x_n)$ is a Cauchy series too.

Currently, I would love to ask the adhering to inquiry:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the adhering to building: for every single Cauchy series $(x_n)$ in $(0,1)$, $(g(x_n))$ is additionally a Cauchy series. Confirm that g is evenly continual on $(0,1)$.

Just how do we deal with doing it?

9
2022-06-07 14:34:33
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Answers: 4

Just a statement: a bottom line is that the open period $(0,1)$ is not full as a statistics room. Intend you have a function $f: X \rightarrow \mathbb{R}$ where $X$ is full . After that $f$ is Cauchy - continual - - i.e., if $\{x_n\}$ is a Cauchy series in $X$, after that $\{f(x_n)\}$ is Cauchy in $\mathbb{R}$ - - iff it is continual. To see that this remains in basic much weak than being evenly continual, take into consideration the instance $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto x^2$.

In the various other instructions, if $X$ is insufficient, yet with portable conclusion, after that every Cauchy continual function $f: X \rightarrow \mathbb{R}$ includes a continual function $\overline{f}: \overline{X} \rightarrow \mathbb{R}$ on the conclusion $\overline{X}$. Given that $\overline{X}$ is portable, $\overline{f}$ is evenly continual, therefore so is its constraint $f$. This last scenario gets in the inquiry asked by the OP.

5
2022-06-07 15:07:50
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You can additionally confirm it by opposition.

Intend that $f$ is not evenly continual. After that there exists an $\epsilon >0$ to make sure that for each and every $\delta>0$ there exists $x,y \in (0,1)$ with $|x-y| < \delta$ and also $|f(x)-f(y)| \geq \epsilon$.

For each and every $n$ choice $x_n, y_n$ to make sure that $|x_n-y_n| < \frac{1}{n}$ and also $|f(x_n)-f(y_n)| \geq \epsilon$.

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

After that the alternaticng series $x_{l_1}, y_{l_1}, x_{l_2}, y_{l_2},..., x_{l_n}, y_{ln}, ...$ is Cauchy yet

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

1
2022-06-07 15:07:18
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$g$ is plainly continual, you actually just require to bother with what would certainly take place at the endpoints. some series $x_n\to 1$ has $g(x_n)\to g(1)$, in a similar way for absolutely no. so you can specify $g$ on $[0,1]$ where it will certainly be evenly continual (primarily $g$ does not zoom off to infinity at the endpoints, else $g(x_n)$ wouldnt be cauchy)

5
2022-06-07 14:55:34
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You need to confirm is that if $(x_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ and also $(y_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ merge to the very same restriction $x \in [0,1]$, after that $(g(x_n))_{n \ge 0}$ and also $(g(y_n))_{n \ge 0}$ have the very same restriction (you might specify $z_{2n} = x_n$, $z_{2n+1} = y_n$, and also make use of individuality of the restriction). After that you can the reason that $g$ is continual on the portable $[0,1]$ (or can be included such a function), therefore evenly continual.

1
2022-06-07 14:54:32
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