Is $g(x)= \frac{x}{1-\left | x \right |}$ uniformly continuous?

Question

I was asked to prove that the function: $$ g: B \subset \mathbb{R}^n \rightarrow \mathbb{R}^n $$ Is uniformly continuous on $B$, where $B$ is the open ball $B(0,1) \subset \mathbb{R}^n$ y $g(x)= \frac{x}{1-\left | x \right |}$

My attempt

I tried to applied the negation of the definition of uniform continuity $$ \exists \space \epsilon_0 > 0 \space | \space \forall \delta > 0, \exists \space x_0, y_0 \in B $$

with

$$ \left | x_0 - y_0 \right | < \delta $$

such as

$$ \left | f(x_0) - f(y_0) \right | \geq \epsilon_0 $$ So I took $\epsilon_0 = 1/2$ and for any $\delta > 0$ I took $x_0=0$ and $y_0 = \frac{\delta}{1+\delta}e_1$, where $e_1 = (1,0,\dots,0)$ so $x_0, y_0 \in B$ and $\left |x_0 - y_0 \right | < \delta$.

But I just obtain: $\left | f(x_0)-f(y_0) \right | = \delta$ which is not always greater than $1/2$.

Any help would be really appreciated.

Answer 1

HINT:

Take $x_0=\left(1-\frac1n\right)e_1$ and $y_0=\left(1-\frac1{2n}\right)e_1$.

Answer 2

Let us first work with the case $n=1$. There is a straightforward way to reduce the general case to this one. Since uniform continuity is a property that a function has to have on its entire domain of definition, if we find that it is not uniformly continuous on a subset of its domain, it is not uniformly continuous, period. So, let us focus on $f(x) = \frac{x}{1-x}$ for $x > 0$.

The idea now to find an appropriate $\epsilon_0$ is to start with one we suspect might work, and typically with functions with vertical asymptotes like this one, any old value for $\epsilon_0$ will work. We want to show that we can make the difference $\Delta f$ very large even though the difference $\Delta x$ may be small. The tricks I will use below aren't strictly speaking necessary, but they make the work simple at each step, even if they may seem like they come out of thin air if you are new to proofs like this. The spirit is always the same: make $\Delta f$ large for arbitrarily small $\Delta x$.

Let's start with the observation that $1 = \frac{1}{1-x} - f(x)$, so $f(x) = \frac{1}{1-x} - 1$ for $x > 0$. Now, a function $g(x)$ is uniformly continuous if and only if the vertical translations $g(x) - a$ with $a\in \Bbb R$ are uniformly continuous, as you can verify with the definition of uniform continuity. So, we can shift our focus to $F(x) = \frac{1}{1-x}$ for $0 < x < 1$. Applying the symmetry $x\mapsto 1-x$ (or changing variables $u = 1-x$ if you like), we can switch our attention again to the function $G(x) = \frac{1}{x}$ with $0 < x < 1$.

Let $\delta>0$ be arbitrary. Then by the mean value theorem, $\big|G\big(\frac{3\delta}{2}\big)-G(\delta)\big| = |G'(c)|\cdot\frac{\delta}{2}$ for some $\delta < c < \frac{3\delta}{2}$. Let's impose $\delta < 1$ since the length of the domain of our function is $1$, so the condition for uniform continuity is trivially not satisfied for $\delta\ge 1$ because the function $G$ is unbounded. Then since $|G'(c)| = \frac{1}{c^2}$, we have $\frac{1}{\delta^2} > |G'(c)| > \frac{4}{9\delta^2}$. Hence $\big|G\big(\frac{3\delta}{2}\big)-G(\delta)\big| > \frac{4}{9\delta^2}\cdot\frac{\delta}{2} > \frac{2}{9}$. Setting $\epsilon_0 = \frac{2}{9}$ gives us the desired result by proving that $G$ is not uniformly continuous for $0 < x < 1$, hence neither is $f$.