Let $f:(0,\infty)\to(0,\infty)$ be uniformly continuous function, is the following statement true?

Let $$f:(0,\infty)\to(0,\infty)$$ be uniformly continuous function. Does it imply $$\lim_{x\to\infty} {f(x+{1\over x})\over f(x)}=1\;?$$

By uniform continuity , for any $$\epsilon>0$$, $$\exists\delta>0$$ such that $$|f(x)-f(y)|<\epsilon$$ as $$|x-y|<\delta$$.
So, $$|f(x+{1\over x})-f(x)|<\epsilon$$ as $$|{1\over x}|<\delta$$ or $$x>\delta$$.
Thus, $$\left| {f(x+{1\over x})\over f(x)}-1\right|<{\epsilon\over f(x)},\ \forall x>\delta$$.
Now, by definition of range set, $$f$$ is bounded below by $$0$$, but can we get a non zero lower bound for function $$f$$ on $$(\delta,\infty)$$ i.e. can we get a $$M>0$$ such that $$f(x)\ge M\ \forall x>\delta$$? If yes then the proof can be completed easily then.

• Please use \lim instead of lim: $\lim_{x \to nifty} \neq lim_{x \to \infty}.$ Apr 13 '19 at 19:31
• Yes, sorry for typing mistake but can you solve it? Apr 13 '19 at 19:38
• The range of the zero function is not $(0,\infty)$ though.
– Mark
Apr 13 '19 at 19:44
• Ok, then you need to disprove the statement with a counter example Apr 13 '19 at 19:44
• The range of the function is given $(0,\infty)$ Apr 13 '19 at 19:46

The statement is not true. Any bounded, continuous function $$f:(0,\infty) \to (0,\infty)$$ where $$f(x) \to 0$$ as $$x \to \infty$$ is uniformly continuous.

Construct such a continuous function which is piecewise linear and where $$f(x) = 1/n$$ for $$x = n$$ and $$f(x) = 2/n$$ for $$x = n + 1/n$$ where $$n \geqslant 2$$ is an integer. The graph of the function looks like a sequence of declining sawtooth peaks.

Here we have $$f(n+1/n)/f(n) \to 2$$ as $$n \to \infty$$.

Thus, it is not the case that $$f(x +1/x)/f(x) \to 1$$ for $$x \in (0,\infty)$$ tending to $$\infty$$.

I'll leave any remaining details to you.

$$f(x) = \begin{cases} 1/2,& 0 \leqslant x \leqslant 2\\1/n, &x = n, n\geqslant 2 \\ 2/n, & x = n + 1/n, n\geqslant 2\\ x - n + 1/n, & n < x < n +1/n, n \geqslant 2\\ 2/n + (1/n-2/n)(x - n - 1/n)/(1-1/n), &n + 1/n < x < n+1, n \geqslant 2 \end{cases}$$

• Can you please write the function properly, $f(x)=1/n\ \forall x\in(0,n), f(x)=2/n \ \forall x\in[n,n+1/n)$ Do you want to write like this? Apr 14 '19 at 1:19
• And one question more, $f:(0,\infty)\to(0,\infty), f(x)=x^2$, it is bounded near zero but not uniformly continuous. Apr 14 '19 at 1:27
• Bounded and continuous on $(0,\infty)$ along with $\lim_{x \to \infty} f(x) = 0$ are sufficient conditions for uniform continuity on $(0,\infty)$. You don't have a limit of $0$ as $x \to \infty$ for $f(x) = x^2$.
– RRL
Apr 14 '19 at 1:40
• I changed the sentence to say bounded , continuous function with a limit of $0$ as $x \to \infty$. That must be UC. I just said bounded near zero originally, but that is redundant. Obviously $f(x) = 1/x$ is unbounded near zero only but not uniformly continuous.
– RRL
Apr 14 '19 at 1:42
• Are you really unable to draw a graph as I described or write the function explicitly yourself. Take any constant for $f(0)$ with a straight line connecting to $f(2) = 1/2$, another line connecting to $f(2+1/2) = 2/2$, another straight line connecting to $f(3) = 1/3$, etc.
– RRL
Apr 14 '19 at 1:46