Function spaces are central examples in metric space theory. Here the points of the space are functions, and distance measures how far two functions are from each other. The uniform metric measures the largest pointwise difference, while the integral metric measures accumulated difference over an interval. :::definition[Space of Bounded Functions] Let $S$ be a nonempty set. Let $B(S)$ denote the set of all bounded scalar-valued functions on $S$. Thus $f\in B(S)$ if there exists $M>0$ such that $$\begin{equation*} |f(x)|\leq M,\qquad x\in S. \end{equation*}$$ ::: :::definition[Uniform Metric on Bounded Functions] For $f,g\in B(S)$, define $$\begin{equation*} d(f,g)=\sup_{x\in S}|f(x)-g(x)|. \end{equation*}$$ Then $d$ is called the $uniform metric$ or $supremum metric$ on $B(S)$. ::: :::theorem[Metric Property of the Uniform Metric on Bounded Functions] The uniform metric defines a metric on $B(S)$. ::: :::proof Given that $f,g\in B(S)$. To prove that $d(f,g)=\sup_{x\in S}|f(x)-g(x)|$ is a metric. The distance is finite because $f-g$ is bounded. The first three axioms follow from the absolute value and equality of functions. For $f,g,h\in B(S)$, $$\begin{equation*} |f(x)-g(x)|\leq |f(x)-h(x)|+|h(x)-g(x)| \end{equation*}$$ for every $x\in S$. Taking supremum over $x$ gives $d(f,g)\leq d(f,h)+d(h,g)$. Hence $d$ is a metric. $\square$ ::: :::definition[Uniform Metric on Continuous Functions] Let $C[a,b]$ be the set of continuous scalar-valued functions on $[a,b]$. Define $$\begin{equation*} d(f,g)=\sup_{x\in[a,b]}|f(x)-g(x)|. \end{equation*}$$ This is the uniform metric on $C[a,b]$. ::: :::remark If $f,g\in C[a,b]$, then $f-g$ is continuous on the compact interval $[a,b]$. Therefore the supremum in the uniform metric exists and is finite. ::: :::definition[Integral Metric on Continuous Functions] Let $C[a,b]$ be the set of continuous scalar-valued functions on $[a,b]$. Define $$\begin{equation*} d(f,g)=\int_a^b |f(x)-g(x)|\,dx. \end{equation*}$$ Then $d$ is a metric on $C[a,b]$. ::: :::proof Given that $f,g,h\in C[a,b]$. To prove that the integral formula defines a metric. Non-negativity and symmetry follow from the absolute value. If $d(f,g)=0$, then the continuous non-negative function $|f-g|$ has integral zero, so $f=g$. The triangle inequality follows by integrating $$\begin{equation*} |f(x)-g(x)|\leq |f(x)-h(x)|+|h(x)-g(x)|. \end{equation*}$$ Hence the integral formula defines a metric on $C[a,b]$. $\square$ ::: :::solved-problem In $C[0,1]$ with the uniform metric, compute the distance between $x(t)=t$ and $y(t)=t^2$. ::: :::solution Let $x(t)=t$ and $y(t)=t^2$. Then $$\begin{equation*} d(x,y)=\sup_{0\leq t\leq 1}|t-t^2|. \end{equation*}$$ Since $t-t^2\geq 0$ on $[0,1]$, maximize $F(t)=t-t^2$. We have $F'(t)=1-2t$, so the critical point is $t=1/2$. Since $F(0)=0$, $F(1)=0$, and $F(1/2)=1/4$, the distance is $$\begin{equation*} d(x,y)=\frac14. \end{equation*}$$ $\square$ ::: :::exercise [1] Define the uniform metric on $B(S)$. [2] Why is the uniform metric finite on $C[a,b]$? [3] Compute the uniform distance between $f(t)=0$ and $g(t)=t$ on $[0,1]$. ::: :::answer [1] $d(f,g)=\sup_{x\in S}|f(x)-g(x)|$. [2] Continuous functions on compact intervals are bounded. [3] The distance is $1$. ::: :::call-to-action[Continue to Metric Limits] subtitle: Next, use the distance function to define sequence convergence in arbitrary metric spaces. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/limits-of-sequences-in-metric-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metrics-on-sequence-spaces :::