Let f(x) = ln(x) on D = (0, inf). Show that f is not uniform continuous. Here's what I got so far:
Assume that f is uniformly continuous on D. Because f is continuous, for all positive epsilon, there exists delta such that |x-y| < d => |f(x) - f(y)| < e. f(x) - f(y) = ln(x/y).
Not sure how to proceed from here.
On
Let $\epsilon$ be 1, then for all $\delta > 0$, we can find two reals x and y such that |x-y|<$\delta$ but $\ln (x) -\ln(y)>1$. In particular, let x be $\delta$, and y be $\delta /e^ 2$. Then 0 < y < x, so |x-y| < x= $\delta$, but $|\ln(x)-\ln(y)|=\ln(\delta)-\ln(\delta /e^2)= \ln(\delta)-(\ln(\delta)-\ln(e^2))=2 > 1$ Thus $\ln(x)$ is not uniform continuous on $(0,\infty)$.
One can show the so-called sequential criterion for uniform continuity (for $\mathbb{R}$) which states that a function $f:D\rightarrow \mathbb{R}$ is not uniformly continuous on $D$ if and only if there exists two sequences $(x_{n})_{n=1}^{\infty},(y_{n})_{n=1}^{\infty}\subseteq D$ such that $|x_{n}-y_{n}|\to 0$ as $n\to\infty$ but $|f(x_{n})-f(y_{n})|\not\to 0$ as $n\to\infty$.
We will use the above result. Define for each $n\in\mathbb{N}$, $x_{n}=\frac{2}{n}$ and $y_{n}=\frac{1}{n+1}$. Then \begin{align*} \left|x_{n}-y_{n}\right|=\left|\frac{2}{n}-\frac{1}{n+1}\right|=\left|\frac{2n+2-n}{n(n+1)}\right|=\frac{n+2}{n(n+1)}\to 0~\text{ as }n\to\infty, \end{align*}
But \begin{align*} |f(x_{n})-f(y_{n})|&=\left|\ln\left(\frac{2}{n}\right)-\ln\left(\frac{1}{n+1}\right)\right| \\ &=\left|\ln\left(\frac{2(n+1)}{n}\right)\right| \\ &=\ln 2+\ln\left(\frac{n+1}{n}\right) \\ &\to\ln 2\neq 0~\text{ as }n\to\infty. \end{align*} Hence, $f=\ln$ is not uniformly continuous on $D=(0,\infty)$.