Square-summability of sequence of averages 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.)
 A: 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 $N\to\infty$.
A: By Hardy's inequality,$$
\sum_{n = 1}^∞ \left( \frac{1}{n} \sum_{k = 1}^n a_k \right)^2 \leqslant 4 \sum_{n = 1}^∞ a_n^2 < +∞.
$$
