Show that the following function is not uniformly continuous on $[a,b)$

Question

Suppose $a,b \in \mathbb{R}$ s.t. $a<b$ and let $f(x) = \sqrt{\frac{x-a}{b-x}}$. Show that $f$ is not uniformly continuous on $[a,b)$.

Finding an $\varepsilon > 0$ and two sequences $\{x_{n}\}$ and $\{y_{n}\}$ in $[a,b)$ s.t. $\lim(x_{n} - y_{n}) = 0$ and $|f(x_{n}) - f(y_{n})| \geq \varepsilon$ would suffice. However I have no idea on how to construct such sequences. Any hints?

Answer 1

Take $x_n$ such that $f(x_n)=n$. This gives $x_n=\frac {a+bn^{2}}{1+n^{2}}$. Then $x_n \to b$. Take $y_n=x_{n+1}$ and $\epsilon =1$.

A shorter proof can be given if you know that uniformly continuous functions on $[a,b)$ are necessarily bounded. Here $f(x) \to \infty$ as $x \to b-$.

Answer 2

To take a more general view, notice that $f$ is continuous, positive, increasing with $\lim\limits_{x \to b} f(x) = \infty$.

Therefore, taking any $\epsilon \gt 0$, you can build the following sequences by induction:

• $x_1 = (a+b)/2$
• $x_{n+1}$ such that $a \lt x_n \lt x_{n+1} \lt b$ and $f(x_{n+1}) \ge f(x_n) + \epsilon$.
• $y_n = x_{n+1}$ for all $n \in \mathbb N$.

$\{x_n\}, \{y_n\}$ are sequences you're looking for.