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

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#### Chris L T521

##### Well-known member
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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 this form. Here's to another year of graduate POTWs! [HR][/HR]
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:
Chebychev's inequality may come in handy here.

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by girdav and Opalg.

Here's girdav's solution:

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:

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):

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.

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