Let be a continuous function such that
We note that
On the other hand by Cauchy – Schwartz we have that
The result follows.
Given a real matrix show that .
We simply note that
The rest is … history!
Let be a real sequence such that , , and . Prove that
Lemma: Let be a bounded sequence of, say, complex numbers and let be another complex number.
where the latter means “convergence on a set of density ” i.e. there exists a subsequence such that and is dense i.e.
Proof: The proof is omitted because it is too technical.
All we need is that the logarithm is continuous and that as well as are bounded sequences.
Note: We can simplify the conditions to and .
The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.
Prove that there exist infinite many positive real numbers such that the sum
It suffices to prove that is continuous. Indeed,
Thus is continuous. Since is non constant , it follows that is an interval. The result follows.