Let \(X\) denote the subset of \(\mathbb{R}^\omega\) consisting of all sequences \((x_1, x_2, \ldots)\) such that \(\sum x_i^2\) converges. (You may assume the standard facts about infinite series. In case they are not familiar to you, we shall give them in Exercise 11 of the next section.)
Show that if \(\mathbf{x}, \mathbf{y} \in X\), then \(\sum \abs{x_i y_i}\) converges. [Hint: Use (b) of Exercise 9 to show that the partial sums are bounded.]
Let \(c \in \mathbb{R}\). Show that if \(\mathbf{x}, \mathbf{y} \in X\), then so are \(\mathbf{x} + \mathbf{y}\) and \(c\mathbf{x}\).
Show that
\[ d(\mathbf{x}, \mathbf{y}) = \left[ \sum_{i=1}^{\infty} (x_i - y_i)^2 \right]^{1/2} \]
is a well-defined metric on \(X\).
Solution
Let \[ X=\left\{\mathbf x=(x_i):\sum_i x_i^2<\infty\right\}. \]
For each \(n\), the finite-dimensional Cauchy–Schwarz inequality of E-UDRRC(b) gives \[ \sum_{i=1}^n|x_iy_i| \le\left(\sum_{i=1}^n x_i^2\right)^{1/2} \left(\sum_{i=1}^n y_i^2\right)^{1/2} \le \|\mathbf x\|_2\|\mathbf y\|_2. \] The partial sums on the left are increasing and bounded, hence converge. Thus \(\sum|x_iy_i|\) converges.
For scalars \(c\), \[ \sum_i(cx_i)^2=c^2\sum_i x_i^2<\infty. \] Also \[ (x_i+y_i)^2\le2x_i^2+2y_i^2, \] so \[ \sum_i(x_i+y_i)^2\le2\sum_i x_i^2+2\sum_i y_i^2<\infty. \] Hence \(X\) is closed under scalar multiplication and addition.
Part (b) shows \(\mathbf x-\mathbf y\in X\), so \[ d(\mathbf x,\mathbf y)=\left(\sum_i(x_i-y_i)^2\right)^{1/2} \] is finite. Nonnegativity, symmetry, and definiteness are immediate. For the triangle inequality, part (a) gives the infinite Cauchy–Schwarz inequality, so for \(u,v\in X\), \[ \|u+v\|_2^2 =\|u\|_2^2+2\sum_i u_iv_i+\|v\|_2^2 \le(\|u\|_2+\|v\|_2)^2. \] Thus \(\|u+v\|_2\le\|u\|_2+\|v\|_2\). Applying this to \[ u=\mathbf x-\mathbf y,\qquad v=\mathbf y-\mathbf z \] gives the triangle inequality for \(d\). Hence \(d\) is a metric on \(X\).