Why Existence Matters
When optimizing a function on a region, it is natural to ask where its largest and smallest values occur. Before trying to locate those points, there is a more basic question: are there points in the region where those values are actually attained?
A function can have values that approach a finite upper bound without ever reaching it. It can also grow without bound, or behave discontinuously near a point in its domain. The Extreme Value Theorem identifies a useful set of conditions that rules out these failures.
Absolute Extrema and Compact Regions
Let \(D\) be a subset of \(\mathbb{R}^n\), and let \(f:D\to\mathbb{R}\). An absolute maximum of \(f\) on \(D\) is a value \(f(\mathbf{x}_{\max})\) such that \(f(\mathbf{x})\leq f(\mathbf{x}_{\max})\) for every \(\mathbf{x}\in D\). An absolute minimum is a value \(f(\mathbf{x}_{\min})\) such that \(f(\mathbf{x}_{\min})\leq f(\mathbf{x})\) for every \(\mathbf{x}\in D\). The points \(\mathbf{x}_{\max}\) and \(\mathbf{x}_{\min}\) are points where the extrema are attained.
A set is compact if every collection of open sets that covers the set has a finite subcollection that still covers it. In Euclidean space, the Heine–Borel theorem gives a convenient equivalent condition: a subset of \(\mathbb{R}^n\) is compact exactly when it is both closed and bounded. Closed means that the set contains all its limit points; bounded means that it lies inside some ball of finite radius.
Both conditions matter to the guarantee: the function must be continuous on the whole domain, and the domain must be compact. The theorem promises that extrema exist, but it does not specify their values or the points where they occur. Also, its conditions are sufficient, not necessary: a function can attain extrema even if one of the conditions fails.
Why the Theorem Holds
The proof uses two standard facts about compact subsets of Euclidean space. First, every sequence in a compact set has a subsequence that converges to a point of that set. Second, a nonempty set of real numbers that is bounded above has a supremum (a least upper bound). Continuity ensures that limits of function values along a convergent sequence are the function value at the limit point.
Suppose, to the contrary, that the values are unbounded. Then for each positive integer \(k\), choose \(\mathbf{x}_k\in D\) such that \(|f(\mathbf{x}_k)|>k\). Compactness gives a subsequence \(\mathbf{x}_{k_j}\) converging to some \(\mathbf{x}\in D\). Continuity implies \(f(\mathbf{x}_{k_j})\to f(\mathbf{x})\), so those values are bounded. This contradicts \(|f(\mathbf{x}_{k_j})|>k_j\), since \(k_j\to\infty\).
Since the values are bounded and \(D\) is nonempty, the set \(f(D)=\{f(\mathbf{x}):\mathbf{x}\in D\}\) has a finite supremum \(M\). For every \(k\in\mathbb{N}\), the definition of supremum gives a point \(\mathbf{y}_k\in D\) such that \(M-\frac{1}{k}<f(\mathbf{y}_k)\leq M\). Compactness gives a subsequence \(\mathbf{y}_{k_j}\) converging to some \(\mathbf{y}\in D\). As \(k_j\to\infty\), the displayed inequalities imply \(f(\mathbf{y}_{k_j})\to M\). Continuity then gives \(f(\mathbf{y})=M\).
The function \(-f\) is continuous on the same compact set, so it attains a maximum there. If that maximum occurs at \(\mathbf{z}\), then \(f(\mathbf{z})\leq f(\mathbf{x})\) for every \(\mathbf{x}\in D\). Thus \(f\) attains an absolute minimum as well.
This argument explains why the two hypotheses work together. Compactness supplies a convergent subsequence whose limit stays in the domain, while continuity carries the limiting function values to the value at that limit point.
Worked Examples: Applying the Existence Guarantee
Worked Example: A radial function on a closed disk
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 4\}\) and \(f(x,y)=5-x^2-y^2\). The disk is closed and bounded, and \(f\) is a polynomial, hence continuous. The theorem guarantees that both absolute extrema exist. We can identify them directly from the defining inequality of the disk:
Subtracting these quantities from \(5\) reverses their order:
Therefore \(1\leq f(x,y)\leq 5\) throughout \(D\). At \((2,0)\), \(x^2+y^2=4\), so \(f(2,0)=5-4=1\). At \((0,0)\), \(x^2+y^2=0\), so \(f(0,0)=5-0=5\). The absolute minimum is \(1\), attained at \((2,0)\) (and other points on the circle), and the absolute maximum is \(5\), attained at \((0,0)\). Substitution verifies both endpoint values and shows that each bound is attained.
Worked Example: A compact region that is disconnected
Consider \(D=[-2,-1]\cup[1,3]\) and \(f(x)=x^2\). This domain is bounded and closed, even though it consists of two separate intervals. Thus it is compact, and \(f\) is continuous on it. On the first interval, \(-2\leq x\leq-1\), squaring gives \(1\leq x^2\leq4\). On the second, \(1\leq x\leq3\), it gives \(1\leq x^2\leq9\). Together the function values lie in \([1,9]\).
The lower bound is attained at \(-1\) and \(1\), and the upper bound is attained at \(3\). Hence the absolute minimum is \(1\), and the absolute maximum is \(9\). This example also shows that the theorem does not require a region to be connected.
When Continuity Fails
Worked Example: A discontinuous function on a compact interval
Let \(D=[0,1]\), which is compact, and define \(g:D\to\mathbb{R}\) by \(g(0)=0\) and \(g(x)=1/x\) when \(0<x\leq1\). For each positive integer \(k\), the point \(x=1/k\) belongs to \(D\) and
These values grow without bound, so \(g\) has no absolute maximum. It does have an absolute minimum: \(g(0)=0\), while \(g(x)=1/x>0\) for every \(x>0\). Thus the minimum is \(0\), attained at \(0\). The function is not continuous at \(0\): for \(x>0\), its values \(1/x\) become unbounded as \(x\) approaches \(0\). The compactness condition holds, but the continuity condition does not, so the theorem gives no guarantee of a maximum.
When Compactness Fails
Worked Example: A continuous function on a nonclosed interval
Let \(D=[0,1)\) and \(h(x)=x\). The function is continuous, and \(D\) is bounded, but it is not closed: its limit point \(1\) is not in \(D\). Therefore \(D\) is not compact. For every \(x\in D\), \(h(x)=x<1\), so \(1\) is an upper bound that is never attained. Moreover, for each \(k\geq2\), the point \(x_k=1-\frac{1}{k}\) belongs to \(D\), and
Thus the values approach \(1\), but \(h\) has no absolute maximum. Its absolute minimum is \(0\), attained at \(x=0\), since \(0\leq x<1\) for all \(x\in D\). Here continuity holds, but compactness fails because the domain omits a limit point. The example shows why an upper bound alone does not guarantee that the greatest value is attained.
What the Guarantee Does—and Does Not—Say
The Extreme Value Theorem is an existence result. It ensures that the largest and smallest values are present somewhere in the domain when its hypotheses hold. It does not tell us how many points attain either value, whether the points are on the boundary or in the interior, or how to find them.
It is also important not to reverse the theorem. If a domain is not compact or a function is not continuous, extrema may still exist; the theorem simply does not guarantee them. For example, the function \(q(x)=x^2\) on the noncompact domain \(\mathbb{R}\) has an absolute minimum of \(0\) at \(x=0\), though it has no absolute maximum. A failed hypothesis removes the general guarantee, not every possibility of an extremum.
Check Your Understanding
Use the theorem's hypotheses carefully when answering these questions.
- State the two conditions on the domain and function in the Extreme Value Theorem.
- Why does compactness matter in the proof when a sequence of function values approaches a supremum?
- Can a compact region be disconnected and still satisfy the theorem? Explain.
- For \(g(0)=0\) and \(g(x)=1/x\) on \(0<x\leq1\), which extremum exists on \([0,1]\), and which does not?
- For \(h(x)=x\) on \([0,1)\), what value do the function values approach near the missing endpoint, and why is it not attained?