Completeness of Bounded Functions in the Sup Metric

Why the Sup Metric Uses One Bound for Every Point

For real-valued functions, the distance at a single input is just the absolute difference between two real numbers. To compare whole functions, we need one number that controls their difference at every input. The sup metric does this by taking the supremum of all those pointwise differences. Its completeness property says that if a sequence of bounded functions becomes uniformly Cauchy in this distance, then its limit is also a bounded function and the convergence to it is uniform.

Let $X$ be a set. No topology on $X$ is needed. A function $f:X\to\mathbb{R}$ is bounded if there is a real number $M\ge 0$ such that $|f(x)|\le M$ for every $x\in X$. Write $B(X)$ for the set of all bounded real-valued functions on $X$.

Definition: If $X\ne\varnothing$ and $f,g\in B(X)$, their sup distance is $$ d_\infty(f,g)=\sup_{x\in X}|f(x)-g(x)|. $$ A sequence $(f_n)$ is Cauchy in this metric if, for every $\varepsilon>0$, there is $N\in\mathbb{N}$ such that $d_\infty(f_n,f_m)<\varepsilon$ whenever $n,m\ge N$. It converges in this metric to $f$ if, for every $\varepsilon>0$, there is $N\in\mathbb{N}$ such that $d_\infty(f_n,f)<\varepsilon$ whenever $n\ge N$.

The supremum in this definition is finite: if $|f(x)|\le M_f$ and $|g(x)|\le M_g$ for all $x$, then $|f(x)-g(x)|\le M_f+M_g$. The supremum need not be attained at any point. If $X=\varnothing$, there is exactly one function from $X$ to $\mathbb{R}$. We give this one-element set the metric that assigns distance $0$ to its sole pair of functions. The substantive proof below concerns $X\ne\varnothing$.

The Sup Distance Is a Metric

Proposition: For nonempty $X$, $d_\infty$ is a metric on $B(X)$.

Proof. For any $f,g\in B(X)$, each $|f(x)-g(x)|\ge 0$, so $d_\infty(f,g)\ge 0$. Also $|f(x)-g(x)|=|g(x)-f(x)|$ for every $x$, giving $d_\infty(f,g)=d_\infty(g,f)$. If $f=g$, every pointwise difference is $0$, so $d_\infty(f,g)=0$. Conversely, if $d_\infty(f,g)=0$, then $0\le |f(x)-g(x)|\le d_\infty(f,g)=0$ for every $x\in X$; hence $f(x)=g(x)$ for every $x$, and $f=g$. Finally, for $f,g,h\in B(X)$, the triangle inequality in $\mathbb{R}$ gives, for each $x\in X$, \[ |f(x)-h(x)|\le |f(x)-g(x)|+|g(x)-h(x)| \le d_\infty(f,g)+d_\infty(g,h). \] Taking the supremum over $x$ gives $d_\infty(f,h)\le d_\infty(f,g)+d_\infty(g,h)$. These are the metric axioms. $\square$

The key feature is that $d_\infty(f_n,f_m)<\varepsilon$ means $|f_n(x)-f_m(x)|<\varepsilon$ for every $x$, with the same $\varepsilon$ regardless of the input. This uniform control lets us build a pointwise limit and then show that the whole sequence approaches it in the sup metric.

Worked Example: A Sequence on a Two-Point Set

Let $X=\{a,b\}$, and for $n\in\mathbb{N}$ define $f_n(a)=2+1/n$ and $f_n(b)=-1+2/n$. Each $f_n$ is bounded because it has only two values. Find its sup-metric limit and verify convergence.

At $a$, the values tend to $2$; at $b$, they tend to $-1$. Thus the candidate limit is the function $f$ with $f(a)=2$ and $f(b)=-1$. For each $n$, the two pointwise errors are \[ |f_n(a)-f(a)|=|(2+1/n)-2|=1/n \] and \[ |f_n(b)-f(b)|=|(-1+2/n)-(-1)|=2/n. \] The supremum over the two-point set is their maximum, so $d_\infty(f_n,f)=\max\{1/n,2/n\}=2/n$. Since $2/n\to 0$, $f_n\to f$ in the sup metric.

Check: for $n=4$, the values are $f_4(a)=2+1/4=9/4$ and $f_4(b)=-1+2/4=-1/2$. The errors from $f$ are $1/4$ and $1/2$, so $d_\infty(f_4,f)=1/2=2/4$, as the formula gives.

Completeness of the Space of Bounded Functions

Theorem: For every set $X$, $B(X)$ is complete in the sup metric. That is, every Cauchy sequence in $B(X)$ converges in the sup metric to a member of $B(X)$.

Proof. If $X=\varnothing$, $B(X)$ has one element, and every sequence in it is constant and converges to that element. Suppose now that $X\ne\varnothing$, and let $(f_n)$ be Cauchy in $d_\infty$.

Fix any $x\in X$. For every $\varepsilon>0$, the Cauchy condition gives an $N$ such that, whenever $m,n\ge N$, \[ |f_n(x)-f_m(x)|\le d_\infty(f_n,f_m)<\varepsilon. \] Thus $(f_n(x))$ is a Cauchy sequence of real numbers. By completeness of $\mathbb{R}$, it converges. Define $f:X\to\mathbb{R}$ by $f(x)=\lim_{m\to\infty}f_m(x)$, for each $x\in X$.

We first show that $f$ is bounded. Apply the Cauchy condition with tolerance $1$. There is $N\in\mathbb{N}$ such that $d_\infty(f_n,f_N)<1$ whenever $n\ge N$. For every $x\in X$ and $n\ge N$, this gives $|f_n(x)-f_N(x)|<1$. Holding $x$ fixed and taking $n\to\infty$, continuity of absolute value on $\mathbb{R}$ gives \[ |f(x)-f_N(x)|\le 1. \] Because $f_N\in B(X)$, there is $M\ge0$ with $|f_N(x)|\le M$ for all $x$. Therefore, for every $x$, \[ |f(x)|\le |f(x)-f_N(x)|+|f_N(x)|\le 1+M. \] This bound holds across all of $X$, so $f\in B(X)$.

It remains to prove convergence in the sup metric. Let $\varepsilon>0$. Since $(f_n)$ is Cauchy, choose $N$ such that $d_\infty(f_n,f_m)<\varepsilon/2$ whenever $n,m\ge N$. Fix $n\ge N$ and $x\in X$. For every $m\ge N$, \[ |f_n(x)-f_m(x)|<\varepsilon/2. \] Letting $m\to\infty$ gives $|f_n(x)-f(x)|\le\varepsilon/2$. This holds for every $x$, so taking the supremum gives $d_\infty(f_n,f)\le\varepsilon/2<\varepsilon$. Thus $f_n\to f$ in the sup metric, proving completeness. $\square$

The proof has three essential stages: real completeness supplies a limit at each input; a comparison with one fixed function $f_N$ shows that the limit is bounded; and the Cauchy estimate, used with the same tolerance for every input, yields convergence in the sup metric. In particular, the argument does not require any structure on $X$, or any continuity or measurability condition on the functions.

Worked Example: Uniform Convergence on the Unit Interval

Let $X=[0,1]$ and define $f_n(t)=t/n$. Each function is bounded: if $0\le t\le1$, then $0\le f_n(t)\le1/n$. The pointwise candidate limit is $f(t)=0$.

For each $t\in[0,1]$, $|f_n(t)-f(t)|=t/n$. Since $t\le1$, this is at most $1/n$, and at $t=1$ it equals $1/n$. Therefore the supremum is attained at $t=1$, and $d_\infty(f_n,f)=1/n$. This tends to $0$, so the convergence is in the sup metric.

We can also check the Cauchy condition directly. For $m,n\in\mathbb{N}$, \[ |f_n(t)-f_m(t)|=t|1/n-1/m|. \] Its supremum over $0\le t\le1$ is $|1/n-1/m|$, attained at $t=1$. If $m,n\ge N$, this is at most $1/N$. Given $\varepsilon>0$, choosing $N>1/\varepsilon$ makes $1/N<\varepsilon$, so the sequence is Cauchy.

Check: at $n=5$, $d_\infty(f_5,f)=1/5$, since the error at $t=1$ is $1/5$ and every error $t/5$ is at most $1/5$.

Worked Example: Pointwise Convergence Without Sup-Metric Convergence

On $X=[0,1]$, define $g_n(t)=\min\{nt,1\}$. Each $g_n$ is bounded between $0$ and $1$. At $t=0$, $g_n(0)=0$ for every $n$; for each fixed $t>0$, $nt\ge1$ once $n\ge1/t$, so $g_n(t)=1$ thereafter. The pointwise limit is $g(0)=0$ and $g(t)=1$ for $t>0$.

To check whether $(g_n)$ is Cauchy in the sup metric, take $m>n$. For $0\le t\le1/m$, both minimum formulas are in their linear ranges, so \[ |g_m(t)-g_n(t)|=(m-n)t\le(m-n)/m=1-n/m. \] For $1/m\le t\le1/n$, $g_m(t)=1$ and $g_n(t)=nt$, so \[ |g_m(t)-g_n(t)|=1-nt\le1-n/m. \] For $1/n\le t\le1$, both functions equal $1$, so the difference is $0$. At $t=1/m$, the difference is exactly $1-n/m$. Hence $d_\infty(g_m,g_n)=1-n/m$. Taking $m=2n$ gives $d_\infty(g_{2n},g_n)=1/2$ for every $n$, so the sequence is not Cauchy in the sup metric.

Check: the formula gives $d_\infty(g_4,g_2)=1-2/4=1/2$. At $t=1/4$, $g_4(1/4)=1$ and $g_2(1/4)=1/2$, so this value is attained.

A Common Pitfall: Confusing Pointwise and Uniform Control

In the last example, each fixed input eventually has zero error from the pointwise limit. That does not mean the errors are small at all inputs at once. Indeed, for $0<t<1/n$, $g_n(t)=nt$ while $g(t)=1$, and therefore $|g_n(t)-g(t)|=1-nt$. These errors have supremum $1$ over $0<t<1/n$: they are less than $1$, but can be made arbitrarily close to $1$. Thus $d_\infty(g_n,g)=1$ for every $n$, not a number tending to zero.

Wrong calculation: “For every fixed $t$, $g_n(t)\to g(t)$, so $\sup_{t\in[0,1]}|g_n(t)-g(t)|\to0$.” This replaces pointwise convergence with a stronger conclusion without justification. The correction is to calculate or bound the supremum at each $n$; here its value remains $1$. The completeness theorem avoids this mistake because it starts from a Cauchy condition in the sup metric, which already controls all inputs with one shared bound.

The result is useful whenever functions are compared by a worst-case error over their whole domain. It guarantees that a uniformly Cauchy sequence does not converge merely to unrelated pointwise values: its limit is still bounded, and the errors from that limit become uniformly small. This conclusion depends on the sup-metric Cauchy hypothesis; pointwise convergence alone does not supply it.

Check Your Understanding

Use the definition of the sup metric and the completeness proof to answer the following.

Why is $d_\infty(f,g)$ finite whenever $f$ and $g$ are bounded?
In the completeness proof, where is completeness of $\mathbb{R}$ used, and where is the boundedness of a particular $f_N$ used?
For $h_n(t)=t/(n+1)$ on $[0,1]$, find $d_\infty(h_n,0)$ and state its limit as $n\to\infty$.
For the functions $g_n(t)=\min\{nt,1\}$, why does pointwise convergence not imply convergence in the sup metric?
If $X=\varnothing$, why is the space of functions from $X$ to $\mathbb{R}$ complete under the stated convention?

Completeness of Bounded Functions in the Sup Metric

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