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Let $(a_n),(x_n)$ and $(y_n)$ be sequences of real numbers with $x_n,y_n\ge0$, $mx_n\le y_n\le M x_n$ for some $m,M>0$, $|x_n a_n|\le1$, $x_n\to0$ and $\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N x_na_n$ exists. Prove that, $\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N y_n a_n$ exists.

Let $L:=\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N x_na_n$ and $S_N=\frac{1}{N}\sum\limits_{n=1}^N y_n a_n$. If I assume $a_n\ge0$. Then observe that $$m\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N x_na_n\le\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N y_n a_n\le M\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{n=1}^N x_na_n$$

This implies $$mL\le\liminf S_N\le\limsup S_N\le ML$$ I tried to prove $S_N$ to be cauchy, but not getting anything.

Can anyone help me with some idea or hint regarding the problem? Thanks for your help in advance.

DeltaEpsilon
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3 Answers3

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The proposition is false and a counter-example can be constructed as follows. Denote $z_{n}=a_{n}x_{n}$. Since there is no restriction imposed on $(a_{n})$, it is easy to choose $(x_{n})$ and $(a_{n})$ such that $z_{n}=1$. Clearly, $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}z_{n}=1$. Now, we go to construct $(y_{n})$ such that if we define $A_{N}=\frac{1}{N}\sum_{n=1}^{N}y_{n}z_{n}$, then $\liminf A_{N}=0$ while $\limsup_{N}A_{N}=1$. Choose $N_{1}$ such that if we define $y_{k}=0$ for $k=1,\ldots,N_{1}$, $A_{N_{1}}\in(0,\frac{1}{1})$. Choose $N_{2}$ and define $y_{k}=1$ for $N_{1}+1\leq k\leq N_{2}$ such that $A_{N_{2}}\in(1-\frac{1}{2},1)$. Choose $N_{3}>N_{2}$ and define $y_{k}=0$ for $N_{2}+1\leq k\leq N_{3}$ such that $A_{N_{3}}\in(0,\frac{1}{3})$. Choose $N_{4}>N_{3}$ and define $y_{k}=1$ for $N_{3}+1\leq k\leq N_{4}$ such that $A_{N_{4}}\in(1-\frac{1}{4},1)$.

Such construction is possible. For, suppose that $y_{1},\ldots,y_{k}$ have been chosen. If we choose $y_{k+1}=y_{k+2}=\ldots=1$, for $K>k$ sufficiently large, we have that $\frac{1}{K}\sum_{i=1}^{K}(y_{i}z_{i})=\frac{1}{K}\sum_{i=1}^{k}y_{i}z_{i}+\frac{K-k}{K}\rightarrow1$ as $K\rightarrow\infty$. On the other hand, if we choose $y_{k+1}=y_{k+2}=\ldots=0$, for $K>k$ sufficiently large, we have that $\frac{1}{K}\sum_{i=1}^{K}(y_{i}z_{i})=\frac{1}{K}\sum_{i=1}^{k}y_{i}z_{i}\rightarrow0$ as $K\rightarrow\infty$. In this way, we can choose sequence $(y_{n})$ from $\{0,1\}$ such that $\liminf A_{N}=0$ while $\limsup_{N}A_{N}=1$.

Danny Pak-Keung Chan
  • 12,923
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Counter-example for the modified question.

Let $x_{n}=\frac{1}{n}$ and $a_{n}=n$. Denote $z_{n}=x_{n}a_{n}=1$. Let $m,M$ be given such that $0<m<M$. We go to choose a sequence $(y_{n})$ that satisfies $mx_{n}\leq y_{n}\leq Mx_{n}$, $\liminf_{N}A_{N}=m$ while $\limsup_{N}A_{N}=M$, where $A_{N}=\frac{1}{N}\sum_{n=1}^{N}a_{n}y_{n}$.

Choose $N_{1}$ and define $y_k=m\cdot x_{k}$ for $k=1,\ldots,N_{1}$ such that $A_{N_{1}}\in[m,m+\frac{M-m}{1})$. Choose $N_{2}>N_{1}$ and define $y_k=M\cdot x_{k}$ for $N_{1}+1\leq k\leq N_{2}$ such that $A_{N_{2}}\in(M-\frac{M-m}{2},M]$. Choose $N_{3}>N_{2}$ and define $y_k=m\cdot x_{k}$ for $N_{2}+1\leq k\leq N_{3}$ such that $A_{N_{2}}\in[m,m+\frac{M-m}{3})$. Choose $N_{4}>N_{3}$ and define $y_k=M\cdot x_{k}$ for $N_{3}+1\leq k\leq N_{4}$ such that $A_{N_{4}}\in(M-\frac{M-m}{4},M]$.

The construction is possible (see my previous post, by observing that $a_{n}y_{n}$ is either $m$ or $M$. Observe that $A_N$ is the running average of a sequence of the form $(m,m,\ldots,M,M,\ldots,m,m\ldots)$). It is now clear that $\lim\inf_{N}A_{N}=m$ while $\limsup_{N}A_{N}=M$.

Danny Pak-Keung Chan
  • 12,923
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Let $a_n>0,$ $a_n\to \infty$ and $x_n=a_n^{-1}.$ Let $b_n$ be a sequence satisfying $m\le b_n\le M.$ For $y_n=x_nb_n.$ we get $${1\over n}\sum_{k=1}^na_ky_k={1\over n}\sum_{k=1}^nb_k\quad (*)$$ It is well known that there are bounded sequences whose Cesàro means $(*)$ are divergent see. The restriction $m\le b_n\le M$ is not essential as we may replace $b_n$ with $\alpha b_n+\beta$ for constants $\alpha\neq 0$ and $\beta.$

Ryszard Szwarc
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