Sequence spaces are among the first infinite-dimensional metric spaces in functional analysis. A point is no longer a number or finite vector; it is an infinite scalar sequence. The supremum metric measures the largest coordinate difference, while the $\ell_p$ metric measures a summable aggregate of coordinate differences. :::definition[Space of Bounded Sequences] Let $X$ be the set of all bounded scalar sequences $x=\{x_i\}_{i\geq 1}$. A sequence belongs to $X$ if $$\begin{equation*} \sup_i |x_i|<\infty. \end{equation*}$$ ::: :::definition[Supremum Metric on Bounded Sequences] Let $X$ be the set of bounded scalar sequences. Define $$\begin{equation*} d(x,y)=\sup_i |x_i-y_i|. \end{equation*}$$ Then $d$ is the $supremum metric$ on the space of bounded sequences. ::: :::theorem[Metric Property of the Supremum Metric on Bounded Sequences] The supremum metric defines a metric on the space of all bounded scalar sequences. ::: :::proof Given that $x$ and $y$ are bounded scalar sequences. To prove that $d(x,y)=\sup_i |x_i-y_i|$ is a metric. The distance is finite because the difference of two bounded sequences is bounded. The first three axioms follow from the absolute value and the definition of supremum. For each $i$, $$\begin{equation*} |x_i-y_i|\leq |x_i-z_i|+|z_i-y_i|\leq d(x,z)+d(z,y). \end{equation*}$$ Taking supremum over $i$ gives $d(x,y)\leq d(x,z)+d(z,y)$. Hence $d$ is a metric. $\square$ ::: :::definition[Summable Sequence Space] Let $p\geq 1$. The space $\ell_p$ is the set of all scalar sequences $x=\{x_i\}_{i\geq 1}$ such that $$\begin{equation*} \sum_{i=1}^{\infty}|x_i|^p<\infty. \end{equation*}$$ ::: :::definition[Metric on Summable Sequence Space] Let $p\geq 1$ and let $X=\ell_p$. Define $$\begin{equation*} d_p(x,y)=\left(\sum_{i=1}^{\infty}|x_i-y_i|^p\right)^{1/p}. \end{equation*}$$ Then $d_p$ is the usual metric on $\ell_p$. ::: :::theorem[Metric Property of the Summable Sequence Metric] Let $p\geq 1$. Then $d_p$ is a metric on $\ell_p$. ::: :::proof Given that $p\geq 1$ and $X=\ell_p$. To prove that $d_p$ is a metric. Non-negativity, identity of indiscernibles, and symmetry follow from the absolute value. For finite partial sums, Minkowski's inequality gives the triangle inequality. Passing to the limit over the partial sums gives $$\begin{equation*} d_p(x,y)\leq d_p(x,z)+d_p(z,y). \end{equation*}$$ Hence $d_p$ is a metric. $\square$ ::: :::example When $p=2$, the space $$\begin{equation*} \ell_2=\left\{x=\{x_i\}:\sum_{i=1}^{\infty}|x_i|^2<\infty\right\} \end{equation*}$$ has metric $d_2(x,y)=\left(\sum_{i=1}^{\infty}|x_i-y_i|^2\right)^{1/2}$. ::: :::solved-problem Let $0<p<1$ and define $d(z,w)=\sum_{k=1}^{\infty}|z_k-w_k|^p$ on the space of $p$-summable scalar sequences. Show that $d$ is a metric. ::: :::solution Let $0<p<1$. The first three metric axioms follow directly from the absolute value and non-negativity of the summands. For the triangle inequality, use $$\begin{equation*} (a+b)^p\leq a^p+b^p,\qquad a,b\geq 0. \end{equation*}$$ Since $|z_k-w_k|\leq |z_k-u_k|+|u_k-w_k|$, we obtain $$\begin{equation*} |z_k-w_k|^p\leq |z_k-u_k|^p+|u_k-w_k|^p. \end{equation*}$$ Summing over $k$ gives $d(z,w)\leq d(z,u)+d(u,w)$. Hence $d$ is a metric. $\square$ ::: :::exercise [1] Define the supremum metric on bounded sequences. [2] Define $\ell_p$. [3] Which inequality proves the triangle inequality for the $\ell_p$ metric when $p\geq 1$? ::: :::answer [1] $d(x,y)=\sup_i |x_i-y_i|$. [2] $\ell_p$ consists of scalar sequences satisfying $\sum |x_i|^p<\infty$. [3] Minkowski's inequality. ::: :::call-to-action[Continue to Function Spaces] subtitle: Next, replace sequences by functions and study uniform and integral metrics. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metrics-on-function-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metrics-on-extended-spaces :::