Let be a continous function and . If be differentiable over , prove that there exist such that .
ah, ok, i think f(0)=f(1/2)=0 is the right condition.
i am assuming that f is actually a positive on [0,1] function, since if not then we can essentially take |f| instead of f.
denote . assume for any ,
first consider interval . by the mean value theorem there: (since f(0)=0), where . therefore for . therefore .
same arguments for (using mean value theorem at the point 1/2, where f is equal to zero) show .
finally consider . again, by the mean value theorem: (since f(1/2)=0), where . therefore for . therefore .
combining, we get , contradiction.