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....