Prove the Extreme Value Theorem

From Intermediate Values to Extreme Values

The Intermediate Value Theorem shows that continuity on an interval forces a function to take every value between its endpoint values. A different question is whether the function reaches its largest and smallest values at all. Continuity alone does not guarantee this: a continuous function on an open interval can approach a limiting value without ever attaining it. Compactness supplies the missing condition.

We prove a slightly more general result than the closed-interval version. The proof uses the Sequential Compactness of a Compact Metric Space, established earlier in this course: every sequence in a compact metric space has a subsequence converging to a point of that space. It also uses the Epsilon Characterization of the Supremum. The key idea is to choose points where the function values approach their supremum, then use compactness to obtain a limit point and continuity to identify its function value.

Attainment is stronger than having a supremum or infimum. A supremum is a least upper bound for the set of function values; it need not itself be a function value. The Extreme Value Theorem says that when the domain is nonempty and compact and the function is continuous, both bounds are attained.

A Continuous Function on a Compact Set Is Bounded

Before taking the supremum of the function values, we must know that those values are bounded above. Rather than assume boundedness, we establish it from compactness and continuity. This argument also illustrates a useful contradiction pattern: an unbounded function would produce a sequence of inputs whose function values escape every bound, while compactness gives a convergent subsequence of those inputs.

Proof. Suppose, to the contrary, that \(f(K)\) is unbounded. For each positive integer \(n\), choose \(x_n\in K\) such that \(|f(x_n)|>n\). By the Sequential Compactness of a Compact Metric Space, there is a subsequence \((x_{n_k})\) converging to some \(x\in K\). Continuity of \(f\) at \(x\) implies that \(f(x_{n_k})\to f(x)\).

Every convergent real sequence is bounded, so the sequence \((f(x_{n_k}))\) must be bounded. But \(n_k\geq k\), and therefore \(|f(x_{n_k})|>n_k\geq k\) for every positive integer \(k\). These values are not bounded, a contradiction. Thus \(f(K)\) is bounded. \(\square\)

The point \(x\) in this proof belongs to \(K\), not merely to a larger space. That is essential: continuity identifies the limiting function value at \(x\) only if \(x\) is in the domain. Sequential compactness provides precisely this guarantee.

The Extreme Value Theorem

Proof. The boundedness lemma shows that \(f(K)\) is bounded. It is nonempty because \(K\) is nonempty. Thus \(s=\sup f(K)\) exists as a real number. For each positive integer \(n\), the Epsilon Characterization of the Supremum gives a point \(x_n\in K\) such that \(s-\frac{1}{n}<f(x_n)\leq s\).

Sequential compactness gives a subsequence \((x_{n_k})\) converging to some \(x_{\max}\in K\). By continuity, \(f(x_{n_k})\to f(x_{\max})\). Also, the displayed inequalities imply \(f(x_{n_k})\to s\): the values are at most \(s\), and their distance below \(s\) is less than \(1/n_k\leq1/k\), which tends to zero. By uniqueness of sequence limits, \(f(x_{\max})=s\). Since \(s\) is an upper bound for \(f(K)\), every \(x\in K\) satisfies \(f(x)\leq s=f(x_{\max})\). Hence \(f\) attains its maximum.

To obtain a minimum, define \(g:K\to\mathbb R\) by \(g(x)=-f(x)\). The function \(g\) is continuous because \(f\) is continuous. The maximum part of the proof gives a point \(x_{\min}\in K\) such that \(g(x)\leq g(x_{\min})\) for every \(x\in K\). Substituting \(g=-f\) and reversing the inequality gives \(f(x_{\min})\leq f(x)\) for every \(x\in K\). Thus \(f\) attains its minimum as well. \(\square\)

The proof does not assume that \(f(K)\) is closed. Instead, it constructs a sequence of values approaching the supremum and shows that compactness and continuity force the supremum to be one of the values. This distinction matters: boundedness alone gives a supremum, but does not ensure that the supremum is attained.

Applying the Theorem to Closed Intervals

Proof. The interval \([a,b]\) is closed and bounded, including when \(a=b\). By the Heine-Borel Theorem, it is compact. It is nonempty because \(a\in[a,b]\). Apply the Extreme Value Theorem with \(K=[a,b]\). It gives points \(x_{\min},x_{\max}\in[a,b]\) attaining the minimum and maximum, respectively. Combining the two inequalities gives the stated conclusion. \(\square\)

When \(a=b\), the interval has just one point. Both extremes are attained there, and the conclusion remains valid. Including this case avoids relying on a hidden assumption that the interval has positive length.

Why the Hypotheses Matter

Compactness and continuity have distinct jobs. Compactness supplies convergent subsequences whose limits remain in the domain. Continuity transfers convergence of inputs to convergence of function values. The proof uses both: without compactness, a sequence approaching the supremum might escape the domain; without continuity, a convergent sequence of inputs need not have function values converging to the value at its limit.

For example, \(f(x)=x\) is continuous on the noncompact interval \((0,1)\), but has no maximum there. Every value is less than \(1\), and values can be chosen arbitrarily close to \(1\), so \(\sup f((0,1))=1\); however, \(1\notin(0,1)\). It also has no minimum, since its infimum is \(0\), which is not in the domain. This does not contradict the theorem: the domain is not compact.

Another common error is to confuse boundedness with attainment. A bounded set of values has a supremum and an infimum, but either can fail to be a value of the function. The maximizing-sequence argument is the step that closes this gap. A reliable proof checks that the domain is nonempty and compact, verifies continuity, and then uses compactness to obtain a limit point for a sequence approaching the relevant bound.