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I am reading a text in which the following is claimed to be a well-known result:

Let $(y_j)_{j=1}^\infty$ be a square-summable sequence of real numbers, that is $$ \sum_{j=1}^\infty y_j^2<\infty. $$ Then the sequence of the averages $(n^{-1}\sum_{j=1}^n y_j)_{j=1}^\infty$ is also square-summable, that is $$ \sum_{n=1}^\infty \left(n^{-1}\sum_{j=1}^n y_j\right)^2<\infty. $$

I tried using Cauchy-Schwarz to verify this but without success:

\begin{align*} \sum_{n=1}^\infty \left(n^{-1}\sum_{j=1}^n y_j\right)^2 &= \sum_{n=1}^\infty n^{-2}\left(\sum_{j=1}^n y_j\right)^2 \le \sum_{n=1}^\infty n^{-2}\sum_{j=1}^n y_j^2 \sum_{j=1}^n 1^2 \\ &= \sum_{n=1}^\infty \sum_{j=1}^n \frac{y_j^2}{n} = \sum_{j=1}^\infty y_j^2 \sum_{n=j}^\infty \frac{1}{n} = \infty. \end{align*} (The last equality holds unless all $y_j$ are zero.)

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  • $\begingroup$ You might want to look for Cesaro summation, which is at least related. $\endgroup$ – Paul Sinclair May 24 '18 at 23:45
  • $\begingroup$ This is a result of Hardy's, isn't it? $\endgroup$ – zhw. May 26 '18 at 23:31
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By Hardy's inequality,$$ \sum_{n = 1}^∞ \left( \frac{1}{n} \sum_{k = 1}^n a_k \right)^2 \leqslant 4 \sum_{n = 1}^∞ a_n^2 < +∞. $$

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Step 1: Preliminary.

  • Let $C > 0$ be a constant such that

    $$\sum_{k=n}^{\infty} \frac{1}{k^2} \leq \frac{C}{n} \tag{1}$$

    for all $n \geq 1$. (It is easy to check that such constant $C$ exists. For instance, one may argue by $\sum_{k=n}^{\infty} \frac{1}{k^2} \leq \frac{1}{n^2} + \int_{n}^{\infty} \frac{dx}{x^2} \leq \frac{2}{n} $ to check that $C = 2$ works for $\text{(1)}$.)

Step 2: Computation. Throughout the computation, the following abbreviations will be used:

  • $\text{IOS}$ : Interchaning the order of summations.
  • $\text{C-S}$ : Cauchy-Schwarz inequality.

Now define $(Ty)_n = \frac{1}{n}\sum_{i=1}^{n} y_i$ and assume first that only finitely many terms of $y_n$'s are non-zero. Then

\begin{align*} \sum_{n=1}^{\infty} (T\lvert y\rvert)_n^2 &= \sum_{n=1}^{\infty} \sum_{i=1}^{n}\sum_{j=1}^{n} \frac{\lvert y_i y_j \rvert}{n^2} \\ &= \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=\max\{i,j\}}^{\infty} \frac{\lvert y_i y_j \rvert}{n^2} \tag{IOS} \\ &\leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{C}{\max\{i,j\}} \lvert y_i y_j \rvert \tag{by (1)} \\ &\leq 2C \sum_{i=1}^{\infty} \sum_{j=i}^{\infty} \frac{1}{j} \lvert y_i y_j \rvert \\ &= 2C \sum_{j=1}^{\infty} \lvert y_j \rvert (T \lvert y \rvert)_j \tag{IOS} \\ &\leq 2C \bigg( \sum_{j=1}^{\infty} y_j^2 \bigg)^{1/2} \bigg( \sum_{j=1}^{\infty} (T \lvert y \rvert)_j^2 \bigg)^{1/2} \tag{C-S}. \end{align*}

So it follows that

$$ \bigg( \sum_{n=1}^{\infty} (Ty)_n^2 \bigg)^{1/2} \leq \bigg( \sum_{n=1}^{\infty} (T \lvert y \rvert)_n^2 \bigg)^{1/2} \leq 2C \bigg( \sum_{n=1}^{\infty} y_n^2 \bigg)^{1/2}. \tag{2} $$

The restriction on $y_n$'s is easily removed by applying $\text{(2)}$ to the truncation $\tilde{y}_n = y_n \mathbf{1}_{n \leq N}$ and then applying the monotone/dominated convergence theorem as $N\to\infty$.

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