What Is the Supremum of the Negative of a Function?

If you negate every value of a function, its upper and lower bounds switch roles. In particular, the supremum of the negative function is determined by the infimum of the original function—not, in general, by its supremum.

Let $A$ be a nonempty domain and let $f:A\to\mathbb{R}$. If the range of $f$ is bounded below, then

$$\sup_{x\in A}(-f(x))=-\inf_{x\in A}f(x).$$

This says nothing by itself about whether either bound is achieved. A supremum can exist even when no input gives that value.

Why negation swaps supremum and infimum

Write $S=\{f(x):x\in A\}$. Suppose $S$ is nonempty and bounded below, and let $m=\inf S$. By definition, $m$ is the greatest lower bound of $S$.

Since $m\leq f(x)$ for every $x\in A$, reversing the inequality when multiplying by $-1$ gives

$$-f(x)\leq -m.$$

Thus $-m$ is an upper bound for the negated values $-S=\{-s:s\in S\}$. To see that it is the least upper bound, let $U$ be any upper bound for $-S$. Then $-s\leq U$ for every $s\in S$, so $-U\leq s$ for every $s\in S$. That makes $-U$ a lower bound for $S$. Because $m$ is the greatest lower bound, $-U\leq m$, or equivalently $-m\leq U$. Every upper bound $U$ is therefore at least $-m$, proving that $\sup(-S)=-m$.

The same fact is often written as

$$\sup_{x\in A}(-f(x))=-\inf_{x\in A}f(x), \qquad \inf_{x\in A}(-f(x))=-\sup_{x\in A}f(x),$$

provided the relevant bounds are finite. Negation reverses the order, so it interchanges the roles of infimum and supremum.

Worked example: the supremum is not attained

Take $f(x)=x$ on the open interval $A=(0,1)$. The values of $f$ are positive and can be arbitrarily close to $0$, so

$$\inf_{x\in(0,1)} f(x)=0.$$

The infimum is not a minimum: there is no $x\in(0,1)$ with $f(x)=0$. Negating the function gives $-f(x)=-x$, whose values lie in $(-1,0)$. They can be arbitrarily close to $0$ from below, and no value equals $0$. Therefore,

$$\sup_{x\in(0,1)}(-f(x))=0=-\inf_{x\in(0,1)}f(x),$$

but $-f$ has no maximum on this domain. This illustrates the distinction between a supremum, which is a least upper bound, and a maximum, which must be one of the values.

If instead the domain were $[0,1]$, then $f$ would attain its infimum at $x=0$, and $-f$ would attain its supremum there. Changing an endpoint can change whether the bound is attained without changing the underlying sign rule.

What if the function is unbounded below?

If $f$ is unbounded below, then for every real number $M$ there is some $x$ with $f(x)<-M$. Consequently, $-f(x)>M$, so the negated function is unbounded above. In the extended real numbers, this is written

$$\sup_{x\in A}(-f(x))=+\infty, \qquad \inf_{x\in A}f(x)=-\infty.$$

Thus the identity still makes sense with extended-real values, if those are allowed. If you are working only with real-valued suprema and infima, say instead that neither finite bound exists in this case.

Common mistakes

• Using the supremum of $f$ instead of its infimum. The correct relationship is $\sup(-f)=-\inf(f)$. The supremum of $f$ instead determines $\inf(-f)$.
• Assuming the supremum is a maximum. The equation identifies a bound, not necessarily a function value. The open-interval example has supremum $0$ but no maximum.
• Negating without reversing the inequality. From $m\leq f(x)$, multiplication by $-1$ gives $-m\geq -f(x)$, not $-m\leq -f(x)$.
• Ignoring whether the range is bounded below. A finite supremum of $-f$ requires $f$ to be bounded below. If it is not, $-f$ is unbounded above.
• Overlooking an empty domain. The usual real-valued statement assumes a nonempty set of function values. Bounds of the empty set depend on the conventions being used, so handle that case separately.

A useful shortcut is to identify the set of values first, then negate its bounds: the lower end of the original set becomes the upper end of the negated set, and vice versa.