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Let $f:[0,\infty)\rightarrow R$ be a continuous function. If $\lim_{x\rightarrow+\infty} f(x)$ is finite, show that $f$ is uniformly continuous.

Also, can I change "$\lim_{x\rightarrow+\infty} f(x)$ is finite" to "$f$ is bounded" and get the same conclusion? Thank you!

Jason Ng
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2 Answers2

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Let $\epsilon>0$ be fixed and suppose $\lim_{x\to\infty} f(x) = L \geq 0.$ Then there exists $x_0$ such that $|L-f(x)|<\epsilon/4$ for all $x\geq x_0.$ In particular, for any $x,y\geq x_0$ we have $|f(x)-f(y)|<\epsilon/2.$

Continuous functions on compact domains are uniformly continuous so there is a $\delta>0$ such that $|f(x)-f(y)|<\epsilon/2$ for all $x,y\in [0,x_0].$

Now suppose $x,y\in [0,\infty)$ and $|x-y|<\delta.$ If $x,y$ are both in $[0,x_0]$ or $[x_0,\infty)$ then $|f(x)-f(y)|<\epsilon.$ If there is one in each segment, assume WLOG $x\in [0,x_0], y\in [x_0,\infty).$

Then $|f(x)-f(y)| = |f(x) - f(x_0)| + |f(x_0)-f(y)| < \epsilon/2 + \epsilon/2 = \epsilon$ so $f$ is uniformly continuous.

I gave you a full solution because I share Paul Garrett's philosophy that students should be given good models for solutions to work from, rather than always be challenged to produce them from scratch. What is important to extract from my answer is 1) the idea behind the working and 2) the technique to make the idea rigorous.

The idea was essentially this: Uniformly continuous means that if you pick an $\epsilon>0$ you must be able to find a $\delta>0$ such that points in the domain that are within $\delta$ distance of each other must have images whose difference is less than $\epsilon.$ It's quite a nice picture if you imagine what this means on the graph. Now we know continuous functions on compact domains are uniformly continuous. The limit condition means the function starts to become near a value $L$ so for big $x$ it doesn't change much, so it certainly seems to satisfy the uniform continuity condition there. Lastly, we can patch together the uniform continuity of the two pieces.

For the second part of your question, the answer is no. You may have seen the theorem that if $f$ is differentiable then it is uniformly continuous if and only if $f'$ is bounded. You should try to prove this. Hint: The mean value theorem is important. So to see if your theorem true, maybe check if it's true with the extra condition that $f$ is differentiable. Can you think of a function that is bounded but has unbounded derivative?

Ragib Zaman
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  • Thanks a lot! I forgot the case when $x$ is less than $x_0$ and $y$ is larger than $x_0$. – Jason Ng Mar 30 '13 at 07:45
  • great answwer.i always like such kind of complete answers – dato datuashvili Mar 30 '13 at 07:48
  • I think of $sin(1/x)$ on $(0,+\infty)$. But this cannot serve as a counter example as the domain is required to be closed. – Jason Ng Mar 30 '13 at 07:56
  • Thinking of nice functions to satisfy a criterion can be a fun challenge, but if you just want to show a result an easier way to to create the function we want. Try to draw a bounded and differentiable function that takes increasingly steep dives from e.g 1 to 0. – Ragib Zaman Mar 30 '13 at 08:04
  • @RagibZaman I'm a little late for this thread but I would really like to thank you for your in-depth answer. Offering both a complete solution and the idea and chain-of-thought behind it really is extremely valuable for beginners. – Nhat Jan 15 '14 at 23:00
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hint

let $\epsilon > 0$, $\exists A > 0$ such that $|f(x) - L| < \frac{\epsilon }{2}$ for $x > A$

where $\lim_{x\rightarrow \infty} f(x) =L$

on$[0,A]$ $ f$ is uniformly contionuous

jim
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  • I think I got it! For $x>A$, $f(x)$'s are close to each other since they are all close to $L$. Correct? – Jason Ng Mar 30 '13 at 07:26
  • the difficult bit is when $x \in[0,A]$ $y \in[A,\infty)$. – jim Mar 30 '13 at 07:41
  • http://math.stackexchange.com/questions/290563/prove-that-f-is-uniform-continuous/290577#290577 same question – jim Mar 30 '13 at 07:46