Absolute Maximum and Absolute Minimum

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

# Absolute Maximum and Absolute Minimum

We will now look at the concept of absolute maximum and absolute minimum values of a function $f : A \to \mathbb{R}$. It is important to emphasize that the set $A$ is majorly important in defining whether $f$ has either an absolute maximum, an absolute minimum, both, or neither.

 Definition: If $A \subseteq \mathbb{R}$ and $f : A \to \mathbb{R}$ is a function, then we say that $f$ has an absolute maximum on $A$ if there exists an $x^* \in A$ such that $\forall x \in A$ we have $f(x^*) ≥ f(x)$ and we define $f(x^*)$ as the absolute maximum. Similarly $f$ has an absolutely minimum on $A$ if there exists an $x_* \in A$ such that $\forall x \in A$ we have $f(x_*) ≤ f(x)$ and we define $f(x_*)$ as the absolute minimum.

For example, consider the function $f : [0, 2] \to \mathbb{R}$ defined by $f(x) = x^2$. We note that the function contains both an absolute maximum and absolute maximum obtained from the point $x^* = 2 \in A$ produces $f(x^*) = f(2) = 4$, while the absolute minimum obtained from the point $x_* = 0 \in A$ produces $f(x_*) = f(0) = 0$.

One such example of a function that contains an absolute minimum but no absolute maximum is the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = n$. The absolute minimum is $1$ since $f(1) = 1$, and $1 = f(1) ≤ f(n) = n$ for all $n \in \mathbb{N}$. We note that this function has no absolute maximum though, as if it did, then it would contradict the Archimedean property.

Of course, there are some functions that contain neither an absolute maximum or absolute minimum on their domain. For example, consider the function $f : (0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ as graphed below.

This function contains no absolute maximum on $(0, \infty)$ as $f$ is not bounded above on $A$. Furthermore, $f$ does not contain an absolute minimum, since as $x \to \infty$, $f(x) \to 0$. For example, if we claim $x$ to produce the absolutely minimum $f(x)$, then take $x < x+1$, and so $f(x) = \frac{1}{x} > \frac{1}{x+1} = f(x+1)$ which contradicts the assumption that $f(x)$ was the absolute minimum.

We have suggested thus far that if a function $f$ has an absolute maximum or absolute minimum on $A$, then there is only one such value for the absolute maximum and absolutely minimum. This should seem rather obvious, but we will prove it formally.

 Theorem 1: If $f : A \to \mathbb{R}$ is a function that contains an absolute maximum $f(x^*)$ then this value is unique. Similarly if $f$ contains an absolute minimum $f(x_*)$ then this value is unique.
• Proof: Suppose that $f$ has two absolute maxima, that is both $f(x_1^*)$ and $f(x_2^*)$ are absolute maximum values of $f$ on $A$. Since $f(x_1^*)$ is an absolute maximum value of $f$ on $A$ then $\forall x \in A$ we have that $f(x_1^*) ≥ f(x)$. Since $x_2^* \in A$ then $f(x_1^*) ≥ f(x_2^*)$.
• Similarly, since $f(x_2^*)$ is an absolute maximum value of $f$ on $A$ then $\forall x \in A$ we have that $f(x_2^*) ≥ f(x)$. Since $x_1^* \in A$ then $f(x_2^*) ≥ f(x_1^*)$.
• From both of these inequalities we conclude that $f(x_1^*) = f(x_2^*)$.
• Now suppose that $f$ has two absolute minima, that is both $f({x_1}_*)$ and $f({x_2}_*)$ are absolute minimum values of $f$ on $A$. Since $f({x_1}_*)$ is an absolute minimum value of $f$ on $A$ then $\forall x \in A$ we have that $f({x_1}_*) ≤ f(x)$. Since ${x_2}_* \in A$ then $f({x_1}_*) ≤ f({x_2}_*)$.
• Similarly, since $f({x_2}_*)$ is an absolute minimum value of $f$ on $A$ then $\forall x \in A$ we have that $f({x_2}_*) ≤ f(x)$. Since ${x_1}_* \in A$ then $f({x_2}_*) ≤ f({x_1}_*)$.
• From both of these inequalities we conclude that $f({x_1}_*) = f({x_2}_*)$. $\blacksquare$