Let be a differentiable function such that forall . Does it follow that is necessarily constant?
contradicting what we had assumed in the first place. Hence is constant.
Let such that . Prove that:
Examine if the series:
as well as the Hermite Hadamard inequality. Applying the Hermite – Hadamard inequality we have that
Exponentiating we get that
This leads us to the conclusion that the series diverges.
Let be a non zero continuous function on a connected surface . Prove that does not change sign on .
and this is an obscurity. Since is connected there exists a continuous curve such that
Define . This function is obviously continuous. Note that . So by Bolzano’s theorem, there exists a such that . This, in return means that but this is impossible since has a constant sign.
Evaluate the series:
(Putnam Competition , 2016)
Since the double series the series converges absolutely we can interchange the summation. Thus: