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Let $x_n$ be a sequence of real numbers such that $$ \sum_{n=1}^\infty |x_n|^2 = \infty. $$ Then there exists a sequence $y_n$ of complex numbers such that $$ \sum_{n=1}^\infty |y_n|^2 < \infty \text{ and } \sum_{n=1}^\infty y_n x_n = \infty. $$

This problem looks innocent enough but I cannot seem to figure out how to start. I was thinking if $N_n$ is a sequence of integers such that $\sum_{k=1}^{N_n} x_k^2 > n$ then I can set $L_ n := \sum_{k=1}^{N_n} x_k^2$ and $y_n := x_n/L_n$. Then for $n \geq N_n$ I have $L_n < 1/n$. I am not really sure how to go on. I was thinking that I could maybe get something like the harmonic series in the case of the series over the product and something like the series over $1/n^2$ in case of the series over $y_n^2$. How to do this just is not clear to me yet.

Edit 1: Added that $x_n$ should be real and $y_n$ complex and added absolute values.

Context

This problem arose in the last lecture of the measure and integration theory class that I just took. The lecturer really cursory introduced measures derived from weak* limits of Riesz products. One defines the functions $$ \mu_{(a_j)}^{n} = \prod_{j = 1}^{n} (1 + a_j \cos(\lambda_j x)) $$ where $(a_j)$ is a sequence of real numbers such that $|a_j| \leq 1$ and $(\lambda_j)$ is a lacunary sequences, i.e. one such that $\lambda_{i+1}/\lambda_{i} \geq 3$ or more clearly, if a real number $x$ can be written as $$ x = \sum_{j = 1}^{+\infty} \epsilon_j \lambda_j $$ where $\epsilon_j \in \{-1, 0, +1\}$ then there is a unique sequence of $(\epsilon_j)$ which produced $x$ in combination with $\lambda_j$.

Now one can show that the weak* limit of these $\mu_{(a_j)}^{n}$ is a measure, which we will call $\mu_{(a_j)}$, since it still depends on the sequence $(a_j)$.

The problem arose in the proof of the following theorem:

Let $(a_j)$ and $(b_j)$ be two sequences as above and $\mu_{(a_j)}$ and $\mu_{(b_j)}$ the associated measures, then these measures are mutually singular if $\sum |a_j - b_j|^2 = \infty$.

In the proof one gets to the point where the above property is needed, with $x_n = a_n - b_n$. It seems like literature about this topic is mostly papers on a very advanced level that my restricted knowledge of measure theory, functional analysis and algebra does not give me access to...

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  • $\begingroup$ In what context did the problem arise? If you're taking a course in functional analysis, a theorem by Banach and somebody else may be the intended way. $\endgroup$ – Daniel Fischer Oct 25 '16 at 15:54
  • $\begingroup$ What would you choose for $y_n$ if $x_n=\frac1{\sqrt{n}}?$ $\endgroup$ – Mitchell Spector Oct 25 '16 at 16:05
  • $\begingroup$ @DanielFischer I added some context $\endgroup$ – Cyianor Oct 26 '16 at 14:37
  • $\begingroup$ @MitchellSpector I was trying to think of something real but I guess $y_n$ must almost be purely complex in this case. My first thought was to set $y_n = (-1)^n/\sqrt{n}$ but that would show exactly the opposite, that $\sum y_n x_n < \infty$ and $\sum |y_n|^2 = \infty$. To make $\sum |y_n|^2$ converge $y_n$ should have to be something which decays slightly faster than $1/\sqrt{n}$ but if $y_n = 1/n^p$ for some $1/2 < p$ than $\sum y_n x_n$ converges... Am I missing something here? My knowledge of series with complex coefficients might be insufficient here. $\endgroup$ – Cyianor Oct 26 '16 at 14:41
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    $\begingroup$ I've added an almost-explicit construction of such a sequence $(y_n)$, you may find it instructive. $\endgroup$ – Daniel Fischer Oct 26 '16 at 15:26
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The existence of such a sequence $(y_n)$ follows easily from the Banach-Steinhaus theorem. For $k \in \mathbb{N}$, let $\lambda_k \colon \ell^2(\mathbb{N},\mathbb{C}) \to \mathbb{C}$ be given by

$$\lambda_k(z) = \sum_{n = 1}^k x_n\cdot z_n.$$

Clearly $\lambda_k$ is a continuous linear functional on the Banach (even Hilbert) space $\ell^2(\mathbb{N},\mathbb{C})$, with

$$\lVert \lambda_k\rVert^2 = \sum_{n = 1}^k \lvert x_n\rvert^2.$$

By the Banach-Steinhaus theorem, if the set $\{ \lambda_k(z) : k \in\mathbb{N}\}$ were bounded for every $z \in \ell^2(\mathbb{N},\mathbb{C})$, then the family $\{\lambda_k : k \in \mathbb{N}\}$ would bee bounded, i.e. there'd exist a $C \in [0,+\infty)$ with $\lVert\lambda_k\rVert \leqslant C$ for all $k$. But that would imply

$$\sum_{n = 1}^\infty \lvert x_n\rvert^2 = \sup_{k\in \mathbb{N}} \sum_{n = 1}^k \lvert x_n\rvert^2 \leqslant C^2 < +\infty,$$

contradicting the premise.

Hence there is at least one $z\in \ell^2(\mathbb{N},\mathbb{C})$ such that $\{ \lambda_k(z) : k\in \mathbb{N}\}$ is unbounded. Pick such a $z$, and define $y_n = e^{i\varphi_n}\cdot z_n$, where $\varphi_n\in \mathbb{R}$ is chosen so that $x_n y_n \geqslant 0$. Then $\lvert y_n\rvert = \lvert z_n\rvert$ for all $n$, so $y\in \ell^2(\mathbb{N},\mathbb{C})$, and

$$\lambda_k(y) = \sum_{n = 1}^k x_n y_n = \sum_{n = 1}^k \lvert x_n z_n\rvert \geqslant \lvert \lambda_k(z)\rvert,$$

so by the unboundedness of $\{\lambda_k(z) : k \in \mathbb{N}\}$ we have

$$\sum_{n = 1}^\infty x_n y_n = \lim_{k\to\infty} \sum_{n = 1}^k x_n y_n = +\infty.$$


One can also almost explicitly construct such a sequence $(y_n)$, given $(x_n)$. Since

$$\sum_{n = 1}^\infty \lvert x_n\rvert^2 = +\infty,$$

we can find a strictly increasing sequence $(N_k)$ (with $N_0 = 0$) such that

$$s_k :=\sum_{n = N_{k-1} + 1}^{N_k} \lvert x_n\rvert^2 \geqslant 1$$

for all $k$. Then define

$$y_n = \frac{1}{k\cdot \sqrt{s_k}}\overline{x}_n$$

for $N_{k-1} < n \leqslant N_k$. This yields a square-summable sequence:

$$\sum_{n = 1}^\infty \lvert y_n\rvert^2 = \sum_{k = 1}^\infty \sum_{n = N_{k-1}+1}^{N_k} \lvert y_n\rvert^2 = \sum_{k = 1}^\infty \frac{1}{k^2s_k}\sum_{n = N_{k-1}+1}^{N_k} \lvert x_n\rvert^2 = \sum_{k = 1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6},$$

and we have

$$\sum_{n = 1}^\infty x_n y_n = \sum_{k = 1}^\infty \sum_{n = N_{k-1}+1}^{N_k} x_n y_n = \sum_{k = 1}^\infty \frac{1}{k\sqrt{s_k}}\sum_{n = N_{k-1}+1}^{N_k} \lvert x_k\rvert^2 = \sum_{k = 1}^\infty \frac{\sqrt{s_k}}{k} \geqslant \sum_{k = 1}^\infty \frac{1}{k} = +\infty.$$

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  • $\begingroup$ The explicit construction is a great add-on! This is what I tried to attempt originally. I have to remember that trick with the "sub-sequence" of partial sums :) $\endgroup$ – Cyianor Oct 27 '16 at 12:21

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