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Extreme value theorem

The extreme value theorem states that a continuous real-valued function on a nonempty compact domain attains both a maximum and a minimum.

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The extreme value theorem is a theorem of mathematical analysis stating that a continuous real-valued function on a closed, bounded interval attains both an absolute maximum and an absolute minimum. More generally, the conclusion holds on any nonempty compact space. The theorem is also called the Weierstrass extreme value theorem, after Karl Weierstrass. Its essential conclusion is attainment: the extreme values occur at points of the domain, rather than merely being approached. (math.ucdavis.edu)

Statement and interpretation

Let a,ba,b be real numbers with a≤ba\leq b, and let

f:[a,b]⟶Rf:[a,b]\longrightarrow\mathbb R

be continuous. Then there exist points xmin⁡,xmax⁡∈[a,b]x_{\min},x_{\max}\in[a,b] such that

f(xmin⁡)≤f(x)≤f(xmax⁡)for every x∈[a,b].f(x_{\min})\leq f(x)\leq f(x_{\max}) \qquad\text{for every }x\in[a,b].

Thus

min⁡x∈[a,b]f(x)=f(xmin⁡),max⁡x∈[a,b]f(x)=f(xmax⁡).\min_{x\in[a,b]}f(x)=f(x_{\min}), \qquad \max_{x\in[a,b]}f(x)=f(x_{\max}).

When a=ba=b, the domain has one point and the conclusion is immediate. The extrema are absolute, or global, because the inequalities compare values throughout the domain, not just near the extremizing points. (uoe-school-of-mathematics.github.io)

Attainment is stronger than boundedness. A bounded function has finite least upper and greatest lower bounds, called its supremum and infimum, but these need not be values of the function. The theorem guarantees that both bounds belong to the function’s range. It does not assert that the extremizing points are unique. (uoe-school-of-mathematics.github.io)

Compactness and the general form

The general statement is:

If KK is a nonempty compact topological space and f:K→Rf:K\to\mathbb R is continuous, then ff attains a maximum and a minimum on KK.

The governing principle from topology is that the continuous image of a compact space is compact. Consequently, f(K)f(K) is a nonempty compact subset of R\mathbb R. It is therefore closed and bounded and contains its supremum and infimum. Because these numbers belong to f(K)f(K), they are attained by ff. This proves the theorem without requiring an interval or any differentiability. (math.dartmouth.edu)

In finite-dimensional Euclidean space, the Heine–Borel theorem identifies compact sets with sets that are both closed and bounded. Hence a continuous function f:K→Rf:K\to\mathbb R, with nonempty closed and bounded K⊆RnK\subseteq\mathbb R^n, attains both extrema. The set need not be an interval or connected. In a general metric space, however, closedness and boundedness alone do not establish compactness; compactness is the relevant hypothesis. (math.toronto.edu)

A sequential proof on an interval

A standard proof uses the Bolzano–Weierstrass theorem.

First, ff is bounded. Otherwise, points xn∈[a,b]x_n\in[a,b] could be chosen with ∣f(xn)∣≥n|f(x_n)|\geq n. The bounded sequence (xn)(x_n) has a convergent subsequence xnk→cx_{n_k}\to c, and closedness gives c∈[a,b]c\in[a,b]. Continuity would imply

f(xnk)⟶f(c),f(x_{n_k})\longrightarrow f(c),

contradicting the unbounded growth of ∣f(xnk)∣|f(x_{n_k})|. (ocw.mit.edu)

Now let M=sup⁡f([a,b])M=\sup f([a,b]). Choose yn∈[a,b]y_n\in[a,b] so that

M−1n<f(yn)≤M.M-\frac1n<f(y_n)\leq M.

Again, a subsequence converges to some d∈[a,b]d\in[a,b]. Continuity and the defining inequalities give

f(d)=lim⁡k→∞f(ynk)=M.f(d)=\lim_{k\to\infty}f(y_{n_k})=M.

Thus the supremum is attained. Applying the same argument to −f-f proves attainment of the minimum. The proof combines boundedness of the domain, retention of subsequential limits within the domain, and preservation of limits by continuity. (ocw.mit.edu)

Why the hypotheses matter

Removing a hypothesis can invalidate the conclusion:

  • A domain that is not closed: f(x)=xf(x)=x on (0,1)(0,1) is continuous and bounded. Its infimum is 00 and its supremum is 11, but neither is attained. (math.ucdavis.edu)

  • An unbounded domain: on R\mathbb R, the continuous bounded function

    f(x)=1−11+x2f(x)=1-\frac{1}{1+x^2}

    has supremum 11, but never equals 11, so it has no maximum. (uoe-school-of-mathematics.github.io)

  • Discontinuity: define f(0)=0f(0)=0 and f(x)=1/xf(x)=1/x for 0<x≤10<x\leq1. Its domain is compact, but it is discontinuous at 00, unbounded above, and has no maximum. (math.ucdavis.edu)

These examples show that the assumptions cannot simply be omitted from a general guarantee. They are not necessary conditions for every individual function to have extrema: for example, a constant function attains both extrema even on an open interval. (openstax.org)

Applications in calculus and optimization

In calculus, the theorem justifies the closed-interval method for finding absolute extrema. For a continuous function on [a,b][a,b], extrema must occur at endpoints or at interior critical points, where its derivative is zero or does not exist. When there are finitely many such candidates, evaluating the function at all of them and at both endpoints identifies the largest and smallest values. Differentiability helps locate candidates; it is not required by the extreme value theorem itself. (openstax.org)

The theorem also supplies the existence step in the proof of Rolle’s theorem, which in turn yields the mean value theorem. A nonconstant continuous function with equal endpoint values must attain at least one extreme value in the interior; differentiability then forces the derivative to vanish there. (openstax.org)

In mathematical optimization, a continuous objective function on a nonempty compact feasible set has global minimizers and maximizers. This is an existence result, not an algorithm for finding them or a guarantee of uniqueness. In several variables, finding extrema generally requires examining the boundary as well as interior critical points. (math.toronto.edu)

Semicontinuous extension

Continuity can be weakened when only one extremum is required. A real-valued upper semicontinuous function on a nonempty compact space attains a maximum; a lower semicontinuous function attains a minimum. These one-sided forms of semicontinuity exclude, respectively, values near a point exceeding its value by a fixed positive amount, or falling below it by such an amount. A continuous function satisfies both conditions, so the usual theorem follows from the two one-sided results. (arxiv.org)

References

  1. Basic Analysis: Introduction to Real Analysisocw.mit.edu
  2. Continuous Functionsmath.ucdavis.edu
  3. The Extreme Value Theoremmath.toronto.edu
  4. Math 63: Winter 2021, Lecture 12math.dartmouth.edu
  5. Continuous functions: three big theoremsuoe-school-of-mathematics.github.io
  6. Fun with "Analysis I": basic theorems in calculus revisitedarxiv.org
  7. 3 Maxima and Minima — Calculus Volume 1openstax.org
  8. 4 The Mean Value Theorem — Calculus Volume 1openstax.org
  9. 7 Maxima/Minima Problems — Calculus Volume 3openstax.org