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Consider the set $\ell^2$ that contains all sequences of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty|a_n|^2<\infty.$$

For two sequences $x=(a_n)$ and $y=(b_n)$ in $\ell^2$, define $$d(x,y)=\left(\sum_{n=1}^\infty|a_n-b_n|^2\right)^\frac12.$$

Prove:

(a) $(X,d)$ is a metric space.

(b) $(X,d)$ is complete.

So far this is what I've done:

(a)

(i) $d(x,y)\geq0$ obviously, and $d(x,y)=0$ iff $a_n=b_n\Rightarrow x=y$.

(ii) $d(x,y)=\left(\sum_{n=1}^\infty|a_n-b_n|^2\right)^\frac12=\left(\sum_{n=1}^\infty|b_n-a_n|^2\right)^\frac12=d(y,x)$.

(iii) Let $z=(c_n)$. We will show that $d(x,z)\leq d(x,y)+d(y,z)$

Equivalently, $$\left(\sum_{n=1}^\infty|a_n-c_n|^2\right)^\frac12\leq \left(\sum_{n=1}^\infty|a_n-b_n|^2\right)^\frac12+\left(\sum_{n=1}^\infty|b_n-c_n|^2\right)^\frac12.$$

I start by seeing that $$\left(\sum_{n=1}^\infty|a_n-b_n|^2\right)^\frac12\leq \left(\sum_{n=1}^\infty\left(|a_n-b_n|+|b_n-c_n|\right)^2\right)^\frac12.$$

And squaring gives: $$\left(\sum_{n=1}^\infty\left(|a_n-b_n|^2+|b_n-c_n|^2+2|a_n-b_n||b_n-c_n|\right)\right)^\frac12$$

Splitting the sum: $$\left(\sum_{n=1}^\infty\left(|a_n-b_n|^2+|b_n-c_n|^2\right)+2\sum_{n=1}^\infty\left(|a_n-b_n||b_n-c_n|\right)\right)^\frac12.$$

Not sure where to go from here, and no idea where to start on the convergence of an arbitrary Cauchy sequence in $\ell^2$, which is needed to show completeness of this metric.

Silvia Ghinassi
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Aidan
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    Instead of keeping the square root, square both sides first. – 9301293 Nov 16 '15 at 18:52
  • Are you saying instead of starting with using the triangle inequality, I should square the left side to obtain $\sum_{n=1}^\infty|a_n-c_n|^2$ ? – Aidan Nov 16 '15 at 19:07
  • I've seen you put thought into typing the math, to make the parenthesis (or any bracket) the appropriate size, use \left( ... \right) (see my edit). – Silvia Ghinassi Nov 16 '15 at 19:09
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    Concerning completeness, take a look here or follow the links therein. – Alex M. Nov 16 '15 at 19:15
  • @AlexM. Thanks! – Aidan Nov 16 '15 at 19:20
  • $( \big(\sum_{n=1}^\infty|a_n-b_n|^2\big)^\frac{1}{2}+\big(\sum_{n=1}^\infty|b_n-c_n|^2\big)^\frac{1}{2})^2 = \big(\sum_{n=1}^\infty|a_n-b_n|^2\big) + \big(\sum_{n=1}^\infty|b_n-c_n|^2\big) + 2 \big(\sum_{n=1}^\infty|a_n-b_n|^2\big)^{\frac 1/2}\big(\sum_{n=1}^\infty|b_n-c_n|^2\big)^\frac{1}{2})$. Look up Schwarz inequality on infinite series to take it from there. – fleablood Nov 16 '15 at 19:30
  • @Aidan yes. This will be an application of Cauchy Swartz. – 9301293 Nov 16 '15 at 19:30

1 Answers1

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Potentially useful remarks:

1) For some fixed $N \in \Bbb{N}$, notice that $a = (a_n)_{n=1}^N$ and $b = (b_n)_{n=1}^N$ are standard vectors in $\Bbb{R}^N$ and the analogous distance is $d(a,b) = \sqrt{\sum\limits_{n=1}^N |a_n - b_n|^2}$ is the standard Euclidean distance in $\Bbb{R}^N$. Can you show that the triangle inequality in this case holds (say using induction on $N$). If so, then the infinite sum is the limiting case $N \to \infty$.

2) If $\{(a_n^{(k)})\}_{k=1}^{\infty}$ is a Cauchy sequence in $\ell_2$, then for each fixed $N$, the sequence $\{a_N^{(k)}\}_{k=1}^{\infty}$ is a Cauchy sequence in $\Bbb{R}$ by the estimate $|a_N^{(k)} - a_N^{(l)}| \leq \operatorname{dist}(a^{(k)},a^{(l)})$. Therefore, for each fixed $N$, there is some value $a_N \in \Bbb{R}$ such that $a_N = \lim\limits_{k \to \infty} a_N^{(k)}$. Thus, the sequence $(a_N)_{N=1}^{\infty}$ is your candidate point of convergence of the Cauchy sequence.

Tom
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