Problem of the Week #52 - May 27th, 2013

[Chris L T521]

This week marks one year since I started posting graduate level POTW questions. I know that they are not easy and not many members attempt these problems, but I'm glad that I still went ahead and posted questions anyways. At this time, I'd like to thank those of you who took part in answering a few of these POTWs in the last year. Again, if you'd like to propose a question for future POTWs, you can fill out http://www.mathhelpboards.com/forms.php?do=form&fid=1. Here's to another year of graduate POTWs! (Bigsmile)

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Many thanks to TheBigBadBen for this week's problem!

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Problem: Suppose that \(\displaystyle E\subset [0,2\pi]\) is a measurable set and that \(\displaystyle \int_E x^n \cos(x) \,dx=0\) for all integers \(\displaystyle n\geq 0\). Show that \(\displaystyle m(E)=0\)

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Hint:

Spoiler

Chebychev's inequality may come in handy here.

 

[Chris L T521]

This week's question was correctly answered by girdav and Opalg.

Here's girdav's solution:

Spoiler

By Stone-Weierstrass theorem, we have that for each continuous functions $f$, $\int_Ef(x)\cos^2(x)dx=0$. Let $F$ be a closed subset of $E$. We can approximate pointwise the characteristic function of $F$ by continuous functions: take $f_n(x):=\frac{d(x,O_n^c)}{d(x,F)+d(x,O_n^c)}$, where $O_n:=\{x,d(x,F)<1/n\}$. By dominated convergence, we get that $\lambda(F)=0$ and we conclude by inner regularity of Lebesgue measure.

Note that by the Müntz–Szász theorem, we can relax the assumption to $\int_E x^{n_k}\cos xdx=0$ for all $k$, where $n_k$ is an increasing sequence of integers such that $\sum_k\frac 1{n_k}$ is divergent.

Here's Opalg's solution:

Spoiler

Define a function $f$ on $[0,/2\pi]$ as the product of $\cos x$ and the characteristic function of $E$. In other words, $f(x) = \begin{cases}\cos x&(x\in E),\\0&(x\notin E) \end{cases}$. Then $f\in L^2(0,2\pi)$ (because it is measurable and bounded). We are told that $f$ is orthogonal to $x^n$, for all $n\geqslant0$. By linearity, $f$ is orthogonal to every polynomial. But polynomials are dense in $L^2(0,2\pi)$. It follows that $f=0$ in $L^2(0,2\pi)$, and thus $f(x)=0$ almost everywhere in the interval. But $\cos x$ is only zero at two points in the interval, so the characteristic function of $E$ must be zero almost everywhere, which is the same as saying that $m(E) = 0.$

And here is TheBigBadBen's solution (which used the hint I provided):

Spoiler

We have the MacLaurin representation of cosine, which is

\(\displaystyle cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \)

We may state that

\(\displaystyle \int_E cos^2(x) \,dx = \int_E \left( \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\right)cos(x) \,dx =
\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \left(\int_E x^{2n}cos(x)\right) \,dx = 0\)

We note that \(\displaystyle f(x):=cos^2x > 0 \) except on a null set.

Now, Chebyshev's inequality states that \(\displaystyle \int_E f \geq \alpha \hspace{1 mm} m\{f\geq \alpha\}\cap E \) for any \(\displaystyle \alpha\geq 0\).

So, suppose that E were not a null set. Then for some \(\displaystyle \alpha>0\), we'd have \(\displaystyle \alpha \hspace{1 mm} m\{f\geq \alpha\} \cap E > 0\), which would mean that \(\displaystyle \int_E f >0\). This contradicts our previous statement.

Thus, E must be of measure 0.