Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f : \mathbb R \rightarrow \mathbb R$ be a continuous function such that $f(x+1) = f(x)$ or $f(x-1)=f(x)$ or $f(1-x)=f(x)$ . then $f$ attains its supremum? here is what I was able to do: $f(x+1)=f(x)$. let $f(x)= \sin(2*\pi*x)$. thus the condition holds. and we know that in this case its supremum is 1.

$f(x-1)=f(x)$ . I took $f(x)= \cos(2*\pi*x)$. true to the condition. and in this case too supremum is 1.

$f(1-x)=f(x)$ . I took $f(x)= \cos(2*\pi*x)$. thus supremum = 1 in this case.

i have taken only particular examples above.what idea i got is $f$ repeats itself after certain interval as $f(x) = f(x+1)$ and all other conditions suggest us that and trigonometric functions like $\sin x$ and $\cos x$ suggests us that supremum does exist. but how can we prove it in general?? and are there other functions which follow such pattern??sorry for my ignorance on such functions. and can u help in explaining how all functions satisfying any one of the above three conditions must have a supremum?? kindly help...thanks in advance....

share|cite|improve this question
Welcome to MSE! It really helps readability to format questions using MathJax (see FAQ). I updated most of your question as an example. Regards – Amzoti Apr 30 '13 at 4:38
thanks...will keep that in mind...:) – under-root Apr 30 '13 at 5:51
You are very welcome! Regards – Amzoti Apr 30 '13 at 5:52
The first two conditions are equivalent if the domain of $f$ is all of $\Bbb R$. Given that, the third says $f(x+\frac 12)$ is odd. You have that the range of $f(x)$ is the range of $f(x)$ over $[0,1]$, a closed interval. – Ross Millikan Apr 30 '13 at 5:52
Strictly speaking, it says: $g(x)=f(x+.5)$ is odd. – genepeer Apr 30 '13 at 5:55

Let $f(x)=f(x+1)$ for all $x$. Since $f$ is continuous on $[0, 1]$ it attains a supremum in that domain. Let it be $f(c)$. Suppose that $f(c)$ was not the supremum on $(-\infty, \infty)$. Then there exists some $c_1$ such that $f(c_1)>f(c)$ but $c_1 \in [n, n+1]$ for some $n \in \mathbb{Z}$. This is absurd since $f(c+n)=f(c)$ is the supremum in this domain. Therefore, $f$ must attain a supremum, $f(c)$. Similar argument for the other case.


The last case $f(x)=f(1-x)$ does not hold. We can not guarantee anything! For example: $f(x)=(x-.5)^2 \cos(x-.5)$ has neither infimum nor supremum.

share|cite|improve this answer
thanks a lot buddy...lots of edits but worth it...kudos!!! :) – under-root Apr 30 '13 at 5:48

Hint: a bounded, continuous function on $\Bbb{R}$ attains its supremum. Can you show that such a function must be bounded?

Edit: The above statement is clearly false, take for example $f(x)=\arctan(x)$.

Hint 2: A continuous function on a closed and bounded interval attains its supremum. See genepeer's answer below.

share|cite|improve this answer
Is it really so? I can think of plenty of continuous bounded function that only tend to their supremum. – zarathustra Apr 30 '13 at 5:07
Hmmmm yes, good point. Mysteriously upvoted incorrect answer! (Can I downvote myself? ;) ) – icurays1 Apr 30 '13 at 5:08
Your edit saves it. Closed interval is the key. – Ross Millikan Apr 30 '13 at 5:49
@icuraysi : this is a nitpick, but for Hint 2 you surely mean a continuous function on a bounded closed interval. $[0, \infty)$ is a closed interval. Let's encourage the OP to be precise. – Stefan Smith May 1 '13 at 2:33
@StefanSmith Fair enough. Answering questions when low on sleep is a very bad idea! – icurays1 May 1 '13 at 3:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.