On page 157 of this site:
http://arxiv.org/pdf/0710.3570.pdf
the author is proving a specific case of one direction of the Muntz-Szasz theorem.
I do not understand the following 3 claims:
1) For $\lambda > 0$ and $x \in (0,1)$, $\lambda x^{\lambda}(1-x) <1$.
Why? I have tried various methods, including induction, and got nowhere.
2) The above claim implies (apparently) for some fixed positive integer $q$:
$||Q_n||_{C[0,1]} \leq |1-{q \over \lambda_{n}}|||Q_{n-1}||_{C[0,1]}$.
See the article for the definition of $Q_n$. It's on page 156 at the beginning of the section.
Why does this follow from my first claim?
3) Why is the last line of the proof true?
Specifically, as $\lambda_n$ goes to $\infty$ by assumption in the proof, how does the last product go to $0$?
I'd suggest quickly reading the section if you're confused by my question. It's very simple up to this point.
Thanks!
This answers #1.
Fix $x \in (0,1)$ and set $f(\lambda) = \ln(\lambda x^\lambda (1-x)) = \ln \lambda + \lambda \ln x + \ln(1-x)$. Then $f'(\lambda) = \frac{1}{\lambda} + \ln x$, so if $\lambda_0 = -\frac{1}{\ln x}$, we have that $f$ is increasing on $(0, \lambda_0)$, decreasing on $(\lambda_0, \infty)$, and so achieves its global maximum at $\lambda = \lambda_0$.
Now $f(\lambda_0) = -\ln(-\ln x) - 1 + \ln(1-x)$. But by the elementary inequality $-\ln x > 1-x$ for $x \in (0,1)$, and the fact that $\ln$ is an increasing function, we have $\ln(-\ln x) > \ln(1-x)$. Hence $f(\lambda_0) < -1$.
We conclude that $f < -1$, which is to say $$\lambda x^\lambda (1-x) < \frac{1}{e} < 1.$$