The Supremum Norm

What the Supremum Norm Measures

For a fixed nonempty domain $E$, the space $B(E)$ collects all bounded real-valued functions on $E$. The supremum norm assigns a nonnegative number to each such function by measuring the smallest uniform bound on its absolute values. The Supremum Norm Is a Norm Theorem established earlier in this course verifies the norm axioms. Here we focus on what the supremum in that definition means, especially when no function value actually reaches it.

Definition: Let $E$ be nonempty and $f\in B(E)$. The supremum norm of $f$ is $\|f\|_\infty=\sup_{x\in E}|f(x)|$. It is the supremum of the set of absolute function values, not necessarily a value attained by $f$.

Because $f$ is bounded, the nonempty set $\{|f(x)|:x\in E\}$ has a finite upper bound. Its supremum is therefore a finite nonnegative real number. If some $x_0\in E$ satisfies $|f(x_0)|=\|f\|_\infty$, then the supremum is a maximum. But boundedness does not ensure such a point exists: the domain may omit a boundary point where values approach their limiting upper bound.

The Least Uniform Bound

The supremum norm can be characterized without first asking whether any value is largest. Consider all nonnegative constants that bound $|f(x)|$ at every point of $E$. The supremum norm is the least such constant. This characterization is useful in estimates: to prove $\|f\|_\infty\leq M$, it is enough to verify $|f(x)|\leq M$ for every $x\in E$.

Theorem (The Supremum Norm Is the Least Uniform Bound): Let $E$ be nonempty and $f\in B(E)$. Then $\|f\|_\infty$ is the least member of the set $\{M\geq0: |f(x)|\leq M\text{ for every }x\in E\}$.

Proof. Write $S=\sup_{x\in E}|f(x)|$. By the definition of supremum, $S$ is an upper bound for all the values $|f(x)|$. Thus $|f(x)|\leq S$ for every $x\in E$, so $S$ itself is one of the nonnegative uniform bounds in the displayed set. If $M\geq0$ is any other member of that set, then $M$ is an upper bound for $\{|f(x)|:x\in E\}$. Since $S$ is the least upper bound, $S\leq M$. Therefore $S$ is the least uniform bound, as claimed. $\square$

A related fact describes how to find values close to the supremum, even when it is not attained. For every positive tolerance, some value must lie within that tolerance below the supremum. Otherwise, a smaller number would already be an upper bound.

Proposition (Approximation to the Supremum): Let $E$ be nonempty, $f\in B(E)$, and $S=\|f\|_\infty$. For every $\varepsilon>0$, there is an $x\in E$ such that $|f(x)|>S-\varepsilon$.

Proof. Suppose, to the contrary, that for some $\varepsilon>0$ every $x\in E$ satisfies $|f(x)|\leq S-\varepsilon$. Then $S-\varepsilon$ is an upper bound for the set of absolute function values. This contradicts the fact that $S$ is its least upper bound, since $S-\varepsilon<S$. Hence there must be an $x\in E$ with $|f(x)|>S-\varepsilon$. $\square$

This proposition does not claim that $|f(x)|=S$ for some point. It says that values can be found arbitrarily close to $S$, which is exactly what distinguishes a supremum from a smaller upper bound. If a maximum does exist, the proposition still holds: a point attaining $S$ works for every $\varepsilon>0$.

Worked Example: A Supremum That Is Not Attained

Define $f:[0,\infty)\to\mathbb{R}$ by $f(x)=x/(1+x)$. For each $x\geq0$, the denominator is positive and $x<1+x$, so $$ 0\leq f(x)=\frac{x}{1+x}<1. $$ Thus $1$ is an upper bound. To show it is the least upper bound, take any $\varepsilon$ with $0<\varepsilon<1$ and set $x=1/\varepsilon$. Then $$ f(1/\varepsilon)=\frac{1/\varepsilon}{1+1/\varepsilon} =\frac{1}{1+\varepsilon}. $$ Moreover, $1/(1+\varepsilon)>1-\varepsilon$, because multiplying by $1+\varepsilon>0$ reduces this inequality to $1>1-\varepsilon^2$. Therefore values of $f$ lie within every positive tolerance of $1$, and $\|f\|_\infty=1$. Yet $f(x)<1$ for every $x\geq0$, so this supremum is not attained.

Worked Example: A Supremum That Is a Maximum

Let $E=[-1,2]$ and $g(x)=3-x$. Since $x\in[-1,2]$, we have $1\leq3-x\leq4$, so $|g(x)|=3-x\leq4$. At $x=-1$, $$ |g(-1)|=|3-(-1)|=4. $$ Consequently, $4$ is both an upper bound and an attained function value, and $\|g\|_\infty=4$. This time the supremum is a maximum.

Norm Balls and Uniform Pointwise Control

The supremum norm also measures the distance between two bounded functions: their distance is $\|f-g\|_\infty$. A bound on this distance controls the difference at every point. In particular, the inequality $\|f-g\|_\infty\leq r$ corresponds exactly to the pointwise inequalities $|f(x)-g(x)|\leq r$ holding throughout $E$. The strict version needs more care, because a family of strict pointwise inequalities need not have a uniform margin.

Proof. Set $h=f-g$, which is bounded by closure of $B(E)$ under subtraction. If $\|h\|_\infty\leq r$, then every $|h(x)|$ is at most its supremum, so $|h(x)|\leq r$ for every $x$. Conversely, if every $|h(x)|\leq r$, then $r$ is an upper bound for the absolute values, and the least-upper-bound property gives $\|h\|_\infty\leq r$. This proves the closed-ball equivalence, including $r=0$.

Now suppose $r>0$. If $\|h\|_\infty<r$, choose $\eta=(r-\|h\|_\infty)/2$. Then $\eta>0$, and $$ |h(x)|\leq\|h\|_\infty <\frac{r+\|h\|_\infty}{2} =r-\eta $$ for every $x\in E$, so in particular $|h(x)|\leq r-\eta$. Conversely, if such an $\eta>0$ exists, then $r-\eta$ is an upper bound for all $|h(x)|$. Hence $\|h\|_\infty\leq r-\eta<r$. This proves the strict-ball equivalence. $\square$

The number $\eta$ in the strict-ball statement is a uniform margin: all pointwise differences stay at least that far inside the radius. Merely knowing that each difference is strictly less than $r$ does not supply a common positive margin. This distinction matters whenever a supremum is approached but not attained.

Worked Example: Strict Pointwise Bounds Without a Strict Norm Bound

Let $E=[0,1)$, and define $h(x)=x$. For every $x\in E$, $0\leq|h(x)|<1$. Nevertheless, $\|h\|_\infty=1$. Indeed, $1$ is an upper bound, and for every $0<\varepsilon<1$, the point $x=1-\varepsilon/2$ belongs to $E$ and satisfies $$ |h(x)|=1-\frac{\varepsilon}{2}>1-\varepsilon. $$ Thus the supremum is $1$, not a smaller number. In particular, $h$ satisfies the strict pointwise inequalities $|h(x)|<1$ everywhere, but it does not satisfy $\|h\|_\infty<1$. There is no positive uniform margin from $1$.

Comparing Two Functions

To compare two functions, apply the same reasoning to their difference. The resulting norm is the smallest nonnegative constant that bounds their pointwise discrepancies. This interpretation is the numerical form of uniform control: one number works simultaneously for every input.

Worked Example: Distance Between Two Functions on an Open Interval

Let $E=(0,2)$, $u(x)=x$, and $v(x)=1$. Their pointwise difference is $u(x)-v(x)=x-1$. Since $0<x<2$, we have $-1<x-1<1$, so $|u(x)-v(x)|<1$ everywhere. Values approach $1$ as $x$ approaches $2$, and approach $1$ in absolute value as $x$ approaches $0$. More explicitly, for any $0<\varepsilon<1$, choosing $x=2-\varepsilon/2$ gives $|u(x)-v(x)|=1-\varepsilon/2>1-\varepsilon$. Therefore $$ \|u-v\|_\infty=\sup_{x\in(0,2)}|x-1|=1. $$ The distance is $1$, although no point in this domain has discrepancy equal to $1$. The open endpoints explain why the supremum is not attained.

A reliable way to establish an exact supremum is to separate the task into two parts. First, prove an upper bound that holds everywhere. Second, show that no smaller number is an upper bound, either by finding a point where the upper bound is attained or by producing values arbitrarily close to it. For an estimate that only needs $\|f\|_\infty\leq M$, the first part is enough; for an exact value, the second part cannot be omitted.

Establish a uniform upper bound.
Verify $|f(x)|\leq M$ for every $x\in E$. This gives $\|f\|_\infty\leq M$.

Test whether the bound is sharp.
Find a point with $|f(x)|=M$, or show that for every $\varepsilon>0$ some value exceeds $M-\varepsilon$.

State attainment separately.
If a point realizes the supremum, it is a maximum. If values only approach it, report the supremum without claiming a maximum.

The supremum norm is therefore more than a convenient notation for the largest function value. It records the least uniform bound, whether or not any input realizes that bound. Its associated distance measures the worst possible discrepancy across the domain, and a strict distance bound means that this discrepancy stays uniformly inside the stated radius.

Takeaway: For $f\in B(E)$, $\|f\|_\infty$ is the least nonnegative constant bounding $|f(x)|$ on all of $E$. It may be approached without being attained. A strict norm bound requires a uniform margin, not merely strict pointwise inequalities.

Check Your Understanding

Use the least-bound and norm-ball characterizations to answer these questions.

Why is the supremum norm itself a valid uniform bound for the absolute values of a bounded function?
What does the Approximation to the Supremum Proposition guarantee when the supremum is not attained?
For $f(x)=x/(1+x)$ on $[0,\infty)$, why is the supremum $1$ even though no value equals $1$?
What pointwise condition is equivalent to $\|f-g\|_\infty\leq r$?
Why do the inequalities $|h(x)|<1$ for every $x\in[0,1)$ fail to imply $\|h\|_\infty<1$ when $h(x)=x$?

What You'll Learn