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The question reads as follows;

$$\text{Let } X \text{ be the subset of } \mathbb R^\omega \text{ consisting of all sequences } (x_1, x_2,\dots) \text{ such that } \sum_{i=1}^\infty x_i^2 \text{ converges. Show that } d(x,y)=[\sum_{i=1}^\infty(x_i-y_i)^2]^\frac{1}{2}\ \text{ is a well defined metric on } X$$

Some relevant results; if $x,y \in X$ then $\sum |x_i y_i|$ converges, $x+y \in X$, $cx \in X$ for $c \in \mathbb R$ and $||x+y|| \leq ||x|| + ||y||$ for the euclidean norm $|| \ ||$.

My attempt:

$d$ is symmetric and $d(x,y) \geq 0$ since $(x_i - y_i)^2 \geq 0, \text{ for all } i$ and equality is reached at $x_i = y_i, \text{ for all } i$. However, for the triangle inequality I do not have the Minkowski inequality at my disposal. After some manipulation I get $[d(x,y)]^2 \leq \sum_{i=1}^\infty (|x_i - z_i| + |z_i - y_i|)^2$ for $x,y,z \in X$. However, it seems I've overshot the evaluation. Perhaps we take the limit of the partial sums, but I'm not quite sure of the details. Any tips?

hampster
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1 Answers1

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Define $\langle x, y \rangle = \sum_i x_i y_i$ and $\| x \| = \langle x, x \rangle^{1/2}$. Then show the Cauchy-Schwarz inequality $|\langle x, y \rangle| \leq \|x\| \|y\|$ and the triangle inequality $\|x+y\| \leq \|x\| + \|y\|$ (hint: expand $\|x+y\|^2 = \langle x+y, x+y \rangle$). The triangle inequality for $d(x,y) = \|x-y\|$ follows.

md2perpe
  • 26,770