I have the following problem involving the set $Y$ of infinite sequences that absolutely converge such that,

$$\sum_{i=0}^\infty x_i^2 \lt\infty$$

where $x_i$ is the $i$-th term in the infinite sequence $x$, together with the metric, $d_2$, defined on $Y$ as

$$d_2(x,y)=\left(\sum_{i=0}^\infty |x_i-y_i|^2 \right)^{1/2} $$

I am now to show that $d_2(x,y)$ is always finite for sequences in $Y$. Here are the facts that I need to make use of:

  1. $\sum_{i=0}^\infty |x_i|\lt\infty$
  2. For $a,b\in \mathbb R, (a-b)^2\le2(a^2+b^2)$

I'm not sure what I am allowed to do involving the absolute values; am I able to keep the absolute value as it is in the above and use the fact from point 2. directly or should I use the fact that,

$$|x|=\sqrt{x^{2}}\implies |x|^2=x^2$$

to get rid of the absolute value at the start? On that point, I was unsure about "putting it back in" at the end, so to speak. Here is what I have done so far:

$$\left(\sum_{i=0}^\infty |x_i-y_i|^2 \right)^{1/2} \le\left(2\sum_{i=0}^\infty |x_i^2-y_i^2| \right)^{1/2} $$

$$\implies\left(2\sum_{i=0}^\infty |x_i^2-y_i^2| \right)^{1/2}-\left(\sum_{i=0}^\infty |x_i-y_i|^2 \right)^{1/2}\ge 0$$

$$\implies\left(\sum_{i=0}^\infty |x_i+y_i|^2 \right)^{1/2}\ge0$$

$$\implies\left(\sum_{i=0}^\infty |x_i+y_i|^2 \right)^{1/2}=\left(\sum_{i=0}^\infty |x_i|^2 \right)^{1/2}+\left(\sum_{i=0}^\infty |y_i|^2 \right)^{1/2}+\left(\sum_{i=0}^\infty 2|x_i||y_i| \right)^{1/2}$$

Which is true because the $\sqrt{}$ function is monotone increasing. Using the fact of 1. it would follow that each of the terms on the right hand side of the last statement would be $\lt\infty$ and so $d_2(x,y)\lt\infty$ as required.

I know what I have to obtain, and think I got it in the end, although I am not sure about the means to get there.

  • $\begingroup$ The inequality on the 3rd line didn't need a proof since it is obvious. I don't see how you obtain the 4th line from the previous ones, and most importantly I also believe it is wrong (save for some very special sequences). $\endgroup$
    – Alex M.
    Aug 4, 2015 at 18:08

2 Answers 2


You're overthinking this. Let $$\ell^2 = \left\{ \{x_n\} : \sum_{n=0}^\infty |x_n|^2<\infty\right\}. $$ If $x,y\in\ell^2$ then for each $n$, $|x_n-y_n|^2\leqslant 2|x_n|^2 + 2|y_n|^2$, so $$\sum_{n=0}^\infty |x_n-y_n|^2 \leqslant 2\sum_{n=0}^\infty |x_n|^2 + 2\sum_{n=0}^\infty |y_n|^2 <\infty, $$ so that $x-y\in\ell^2$ and $d_2(x,y)=d_2(x-y,0)<\infty$.

  • 1
    $\begingroup$ You were right about the over thinking it - cheers! $\endgroup$ Aug 4, 2015 at 18:25

This is trivial. Note that $d_2 (x,y) = d_2 (x-y,0)$. Also, note that if $x$ and $y$ are square-summable, then so is $x-y$ (i.e. square-summable sequences form a vector space). Then there exist a number $M$ such that $\sum \limits _{i=0} ^\infty (x_i - y_i)^2 = M$, therefore $d_2 (x-y,0) = \sqrt M$.

(To show that the difference $x-y$ is square-summable too, use relationship 2 in your question: $\sum \limits _{i=0} ^\infty (x_i - y_i)^2 \le \sum \limits _{i=0} ^\infty 2 (x_i ^2 + y_i ^2) = 2 \big( \sum \limits _{i=0} ^\infty x_i ^2 + \sum \limits _{i=0} ^\infty y_i ^2 \big)$ and each series is finite by assumption.)


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