Prove $f(x)$ is not uniformly continuous on $(0,1)$.

Question

Explain or prove why $f(x)=\ln(x)$ is not uniformly continuous on $(0,1)$.

By definition $f$ is uniformly continuous if for all $\varepsilon >0$, there exists a $\delta > 0$ such that when $|x-y|< \delta$ we get $|f(x)-f(y)| < \varepsilon$.

I am not sure where to start. Any suggestions?

Answer 1

Take for example

$$\begin{cases}x_n:=\frac1n\in(0,\infty)\\{}\\y_n:=\frac2n\in(0,\infty)\end{cases}\implies\left|\frac1n-\frac2n\right|=\frac1n\xrightarrow[n\to\infty]{}0$$

yet

$$\left|\log\frac1n-\log\frac2n\right|=\log2\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0\implies f(x)$$

can't be uniformly continuous in $\;(0,\infty)\;$ .

Answer 2

Let $x_n=e^{-n}$. Then $\vert x_{n+1}-x_n\vert\to0$, but $\vert f(x_{n+1})-f(x_n)\vert=1$.

This means that for any $\delta>0$, there are points within $\delta$ of one another, whose images under $f$ are separated by at least $1$.

As such, $f$ cannot be uniformly continuous.

Answer 3

There is a picture you should have in your head. Around each point in the graph of the function, construct a rectangle of half-width $\delta$ and half-height $\varepsilon$. (That is, the rectangle $(x-\delta, x+\delta) \times (y-\varepsilon, y+\varepsilon)$ for $(x,y)$ on the graph of $f$.) Various forms of continuity ask about dimensions of the rectangles so the function only exits the left and right sides of the rectangles, not the top and bottom sides. The function is continuous if for a fixed height across all the rectangles, each rectangle can have a width where the function does not exit the top or bottom of the rectangle. The function is uniformly continuous if for each height there is a width for all the rectangles (making all the rectangles congruent) where the function does not exit the top or bottom of any rectangle.

Is there a region on the graph of the logarithm function where it might be difficult to draw rectangles that the function does not exit through the bottom? As a hint: Notice that $x_n = 1/n$ is a Cauchy sequence in $(0,1)$ (for $n = 2, 3, 4, \dots$). We see $x_{n+1} - x_n \rightarrow 0$. What is $|\ln(x_{n+1}) - \ln(x_n)|$ doing? Given that, can you suggest a different sequence whose logs of ratios does not go to zero?

How about $x_n = \mathrm{e}^{-n}$? Then $\left| \ln \frac{\mathrm{e}^{-n-1}}{\mathrm{e}^{-n}} \right|$ does what you want.