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….
Best Answer
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.
Edit:
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.