Convergence of Square of Linear Combination of Sequences whose Squares Converge

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Theorem

Let $\sequence {x_i}$ and $\sequence {y_i}$ be real sequences such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ and $\displaystyle \sum_{i \mathop \ge 0} y_i^2$ are convergent.

Let $\lambda, \mu \in \R$ be real numbers.


Then $\displaystyle \sum_{i \mathop \ge 0} \paren {\lambda x_i + \mu y_i}^2$ is convergent.


Proof

Let $n \in \N$.

Then:

$\displaystyle \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2 = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \lambda \mu \sum_{i \mathop = 1}^n x_i y_i$

By Cauchy's Inequality:

$\displaystyle \sum_{i \mathop = 1}^n x_i y_i \le \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2}$

Hence:

\(\displaystyle \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2\) \(\le\) \(\displaystyle \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \sum_{i \mathop = 1}^n x_i y_i\)
\(\displaystyle \) \(\le\) \(\displaystyle \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\size \lambda \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2} }^2\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {\size \lambda \paren {\sum_{i \mathop \ge 0} x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop \ge 0} y_i^2}^{\frac 1 2} }^2\)

Thus the sequence of partial sums $\displaystyle \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is bounded above.

We also have that $\displaystyle \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is also increasing.

So by the Monotone Convergence Theorem, $\displaystyle \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is convergent.

$\blacksquare$


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