Exercise 20.10

Proof. First, denote the partial sums of the series $\sum <!--\; FUNCTION\; APPLICATION\; -->\left|x_i{y}_i\right|$ by

for any $n\in {\mathbb{Z}}_{\text{+}}$. Clearly then each term in the sum is non-negative so that the sequence of partial sums is non-decreasing, which is to say that $s_{n+1}\ge s_n$ for every $n\in {\mathbb{Z}}_{\text{+}}$. Then, to show that the series converges, it suffices to show that the partial sums are bounded since convergence then follows from what will be shown in Exercise 21.11 part (a). To this end, we show that the sequence $\left(s_1,s_2,\dots \right)$ is a bounded above.

So, first, let us define the real numbers

which we know are well defined since $x,y\in X$. Also, for any $n\in {\mathbb{Z}}_{\text{+}}$, define the truncated sequences $x_n=\left(x_1,\dots ,x_n\right)$ and $y_n=\left(y_1,\dots ,y_n\right)$. We know from Lemma 20.8.4 that the partial series ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=n}^{\infty }{x}_i^2$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=n}^{\infty }{y}_i^2$ are non-negative for any $n\in {\mathbb{Z}}_{\text{+}}$. From this and Definition 20.8.1 we have that

for any $n\in {\mathbb{Z}}_{\text{+}}$. It then follows from Corollary 1 that

and similarly $\Vert y_n\Vert \le \Vert y\Vert$ for any $n\in {\mathbb{Z}}_{\text{+}}$. Hence, clearly $\Vert x_n\Vert \Vert y_n\Vert \le \Vert x\Vert \Vert y\Vert$ since norms are always non-negative by Lemma 20.9.1 part (4).

Lastly, we define ${\overline{x}}_n=(\left|x_1\right|,\dots ,\left|x_n\right|)$ and ${\overline{y}}_n=(\left|y_1\right|,\dots ,\left|y_n\right|)$ for any $n\in {\mathbb{Z}}_{\text{+}}$. From this definition, it follows that

and similarly that $\Vert {\overline{y}}_n\Vert =\Vert y_n\Vert$.

Putting all of this together we have by Exercise 20.9 part (b) that

for any $n\in {\mathbb{Z}}_{\text{+}}$. This shows that the sequence of partial sums is bounded by $\Vert x\Vert \Vert y\Vert$, which shows the desired result as previously discussed. □

Proof. First, it is a well-known fact that if a series converges absolutely, then it converges. Hence, since it was shown in part (a) that $\sum <!--\; FUNCTION\; APPLICATION\; -->\left|x_i{y}_i\right|$ converges, we have that $\sum <!--\; FUNCTION\; APPLICATION\; -->x_i{y}_i$ also converges. We of course also know that $\sum <!--\; FUNCTION\; APPLICATION\; -->x_i^2$ and $\sum <!--\; FUNCTION\; APPLICATION\; -->y_i^2$ converge since $x,y\in X$. It then follows from Exercise 21.11 part (b) that

converges to $\sum <!--\; FUNCTION\; APPLICATION\; -->x_i^2+2\sum x_i{y}_i+\sum y_i^2$. Since of course

we have that $x+y\in X$ as desired. We also have that $cx=\left(c{x}_1,c{x}_2,\dots \right)$ by definition, and it again follows from Exercise 21.11 part (b) that

converges to $c^2\sum <!--\; FUNCTION\; APPLICATION\; -->x_i^2$. Hence $cx\in X$ as desired. □

Proof. Suppose that $x,y\in X$. It then follows from part (b) that $-y=(-1)y\in X$. Then also $x-y=x+(-y)\in X$ as well, again by what was shown in part (b). Since we clearly have

it therefore follows that $\sum <!--\; FUNCTION\; APPLICATION\; -->{(x_i-y_i)}^2$ converges since $x-y\in X$. Hence the function

is well-defined.

To show that $d$ is a metric, first it is obvious that $d(x,y)\ge 0$ for all $x,y\in X$ since each term is non-negative so that the sequence of partial sums is non-negative and non-decreasing. Also, clearly if $y=x$ then

To show the converse, suppose that $x\ne y$ so that there is an $n\in {\mathbb{Z}}_{\text{+}}$ where $x_n\ne y_n$, and hence $x_n-y_n\ne 0$ and ${(x_n-y_n)}^2>0$. Referencing Definition 20.8.1, we then have

Thus by Corollary 1 we have

so that of course $d(x,y)\ne 0$ as desired. This shows part (1) of the definition of a metric.

Showing part (2) of the definition is even easier:

For part (3) consider any $x,y,z\in X$. Denote the truncated sequences by $x_n=\left(x_1,\dots ,x_n\right)$ for any $n\in {\mathbb{Z}}_{\text{+}}$, and similarly for $y_n$ and $z_n$. Then we have

Since the function $f(x)=x^{1∕2}$ is continuous on the domain $\left\{x\in ℝ\mid x\ge 0\right\}$ by elementary calculus, it follows from Theorem 21.3 in the next section that

Since $x$ and $y$ are arbitrary, we of course also have that

Now, it was shown in Exercise 20.9 part (d) that

for any $n\in {\mathbb{Z}}_{\text{+}}$. It then follows from Lemma 20.8.1 that

as desired. Note that we have used the well-known property of sequences that, if $\left(a_1,a_2,\dots \right)$ and $\left(b_1,b_2,\dots \right)$ are real sequences that converge to $a$ and $b$, respectively, then $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}(a_n+b_n)=a+b$. (This is shown later in Exercise 21.5.) This shows that $d$ has all three of the properties required to be a metric. □