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Extreme value theorem

The Extreme Value Theorem states that if a real-valued function f is continuous on a closed and bounded interval [a, b], then f attains both an absolute maximum value and an absolute minimum value on [a, b].^[1]^[2] This result guarantees the existence of global extrema without specifying their locations, distinguishing it from local analysis techniques like derivatives.^[1]^[3] The theorem is a cornerstone of real analysis and calculus, underpinning the theoretical framework for optimization by ensuring that continuous functions on compact domains are bounded and achieve their bounds.^[4]^[5] It implies that the image of a compact set under a continuous map is compact, which directly follows from the Heine-Borel theorem characterizing compactness in \mathbb{R}^n.^[6] In practice, this allows mathematicians and scientists to confidently seek maxima and minima in problems ranging from physics simulations to economic modeling, where continuity on a finite domain is often assumed.^[7]^[8] Proofs of the theorem typically rely on the completeness of the real numbers and the least upper bound property, often constructing sequences to show attainment of suprema and infima.^[9] For instance, one standard approach assumes the supremum is not achieved and derives a contradiction using the Bolzano-Weierstrass theorem on bounded sequences.^[9] The result extends naturally to higher dimensions: a continuous function on a compact subset of \mathbb{R}^n attains its extrema, making it vital for multivariable calculus and topology.^[10]^[6] The extreme value theorem was first proved by Bernard Bolzano in the 1830s in his unpublished work Function Theory, which appeared in print in 1930; it was independently proved by Karl Weierstrass around 1860.^[11]^[12] Today, it serves as a foundational tool in fields like numerical analysis and machine learning, where ensuring extremal values exist aids algorithm design for constrained optimization.^[8]

Fundamental Concepts

Continuous Functions on Intervals

A function f: D \to \mathbb{R}, where D \subseteq \mathbb{R}, is continuous at a point a \in D if for every \varepsilon > 0, there exists \delta > 0 such that if x \in D and |x - a| < \delta, then |f(x) - f(a)| < \varepsilon.^[13] This \varepsilon-\delta definition, formalized by Karl Weierstrass in the 19th century, captures the intuitive notion that small changes in input near a produce small changes in output.^[14] A function is continuous on an interval I \subseteq \mathbb{R} if it is continuous at every point in I.^[15] On a closed bounded interval [a, b], every pointwise continuous function is uniformly continuous, where for every \varepsilon > 0, there exists \delta > 0 such that for all x, y \in [a, b] with |x - y| < \delta, |f(x) - f(y)| < \varepsilon, independent of the points chosen.^[16] This follows from the Heine-Cantor theorem, which strengthens the local control of continuity to global uniformity over compact domains.^[17] This property prevents pathological behaviors like those on unbounded domains.^[18] A key property of continuous functions on intervals is the intermediate value theorem: if f is continuous on [a, b] and k lies between f(a) and f(b), then there exists c \in [a, b] such that f(c) = k.^[19] First proved by Bernard Bolzano in 1817 and Augustin-Louis Cauchy shortly after, this theorem underscores how continuity ensures the function's image fills all intermediate values without gaps.^[20] Examples of continuous functions on intervals include linear functions f(x) = mx + c, which satisfy the \varepsilon-\delta condition with \delta = \varepsilon / |m| for m \neq 0, and polynomials of any degree, whose continuity follows from the composition and sum of continuous functions.^[15] To illustrate that continuity does not require differentiability, consider the Weierstrass function f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x) with $0 < a < 1 and ab > 1 + \frac{3\pi}{2}, which is continuous everywhere but differentiable nowhere on \mathbb{R}.^[21] This construction, introduced by Karl Weierstrass in 1872, demonstrates the existence of highly irregular yet continuous functions on intervals.^[22]

Compact Sets

In the context of the real numbers \mathbb{R}, a set is called closed if it contains all its limit points. A limit point of a set E \subseteq \mathbb{R} is a point x \in \mathbb{R} such that every neighborhood of x contains at least one point of E distinct from x itself.^[23] Equivalently, a set is closed if its complement is open, where an open set consists entirely of interior points without boundary points included.^[24] A set E \subseteq \mathbb{R} is bounded if there exists a finite interval [a, b] with a < b such that E \subseteq [a, b], meaning all elements of E lie within some interval of finite length.^[25] The Heine-Borel theorem characterizes compactness on the real line by linking it directly to these properties: a subset of \mathbb{R} is compact if and only if it is closed and bounded.^[26] This result, named after Eduard Heine and Émile Borel, establishes that closed and bounded intervals like [a, b] with a \leq b are precisely the compact subsets of \mathbb{R}.^[27] One standard definition of compactness for a subset K \subseteq \mathbb{R} is topological: K is compact if every open cover of K admits a finite subcover. An open cover of K is a collection \{U_\alpha\}_{\alpha \in A} of open sets in \mathbb{R} such that K \subseteq \bigcup_{\alpha \in A} U_\alpha.^[28] For instance, the closed interval [0, 1] is compact because any open cover, such as the collection of open intervals \{(-\frac{1}{n}, 1 + \frac{1}{n})\}_{n=1}^\infty, has a finite subcover (in fact, the single set with n=1 suffices). In contrast, the open interval (0, 1) is not compact; the open cover \{( \frac{1}{n+1}, 1 ) \}_{n=1}^\infty has no finite subcover, as any finite collection misses points near 0.^[29] The Heine-Borel theorem proves that closed and bounded sets satisfy this cover property, while unbounded or non-closed sets do not.^[30] In \mathbb{R}, compactness is equivalent to sequential compactness: a set K \subseteq \mathbb{R} is compact if and only if every sequence in K has a subsequence converging to a point in K.^[31] This equivalence holds because \mathbb{R} is a metric space, where the two notions coincide. The Bolzano-Weierstrass theorem underpins this by stating that every bounded sequence in \mathbb{R} has a convergent subsequence (with limit in \mathbb{R}).^[32] For a closed set, the limit of any convergent subsequence from a bounded sequence lies within the set, thus linking boundedness and closedness to sequential compactness and, via the equivalence, to the full topological compactness required for theorems on continuous functions.

Core Theorems

Boundedness Theorem

The Boundedness Theorem states that if f: K \to \mathbb{R} is a continuous function and K \subset \mathbb{R} is a compact set, then the image f(K) is bounded; that is, there exists a real number M > 0 such that |f(x)| \leq M for all x \in K.^[33]^[34] This result holds because compactness in \mathbb{R} equates to closed and bounded intervals (by the Heine-Borel theorem), and continuity ensures the function cannot escape to infinity without violating its definition on such a restricted domain.^[35] Intuitively, the continuity of f precludes abrupt jumps or oscillations that might produce unbounded values within the finite extent of a compact set K, as the function's behavior is locally controlled and the domain lacks "room" for divergence.^[34] Without continuity, boundedness may fail even on compact domains, but the theorem leverages the uniform control provided by continuity to guarantee a finite range.^[33] A classic counterexample illustrates the necessity of compactness: consider f(x) = \frac{1}{x} on the interval (0,1], which is continuous but unbounded as x approaches 0 from the right, since f(x) \to \infty.^[36] In contrast, restricting to the compact interval [1,2] yields \frac{1}{2} \leq f(x) \leq 1, confirming boundedness.^[35] To sketch the proof, suppose f is unbounded on the compact K; then for each positive integer n, there exists x_n \in K with |f(x_n)| > n. By compactness, the sequence \{x_n\} has a convergent subsequence \{x_{n_k}\} with limit x \in K. Continuity implies f(x_{n_k}) \to f(x), a finite value, contradicting |f(x_{n_k})| > n_k \to \infty. Alternatively, using open covers: for each t \in K, continuity yields an open neighborhood I_t where |f(y) - f(t)| < 1 for y \in I_t \cap K; the collection \{I_t\} covers the compact K, so a finite subcover exists, and bounding f on finitely many points extends to all of K via the Heine-Borel theorem.^[34]^[35]

Extreme Value Theorem

The Extreme Value Theorem states that if f: [a, b] \to \mathbb{R} is continuous on the closed interval [a, b], then there exist points c, d \in [a, b] such that f(c) = \min_{x \in [a, b]} f(x) and f(d) = \max_{x \in [a, b]} f(x).^[11] This result, also known as the Weierstrass extreme value theorem, ensures that continuous functions on compact sets attain both their minimum and maximum values within the domain./03:_The_Graphical_Behavior_of_Functions/3.01:_Extreme_Values) The theorem follows from the combination of boundedness and the attainment property: since f is bounded on [a, b], it has finite infimum m = \inf_{x \in [a, b]} f(x) and supremum M = \sup_{x \in [a, b]} f(x); continuity then guarantees the existence of points where f achieves these bounds./07:_Intermediate_and_Extreme_Values/7.04:_The_Supremum_and_the_Extreme_Value_Theorem) This builds on the boundedness theorem as a prerequisite for the finiteness of m and M./07:_Intermediate_and_Extreme_Values/7.04:_The_Supremum_and_the_Extreme_Value_Theorem) For example, the function f(x) = x^2 on [-1, 1] attains its minimum value of 0 at x = 0 and its maximum value of 1 at x = \pm 1./03:_The_Graphical_Behavior_of_Functions/3.01:_Extreme_Values) In optimization, the theorem provides a foundational guarantee that global maxima and minima exist for continuous objective functions over compact domains, enabling the analysis of extrema without reliance on derivatives.^[37]

Counterexamples and Limitations

Failure Due to Non-Compact Domains

The Extreme Value Theorem requires the domain to be a compact set in \mathbb{R}, which, by the Heine-Borel theorem, means it must be closed and bounded.^[38] Without compactness, continuous functions may fail to attain their extrema, even if bounded, because non-compact domains can admit sequences without convergent subsequences in the domain.^[39] A classic counterexample occurs on an open interval, which lacks closure. Consider the continuous function f(x) = x defined on the domain (0, 1). This function is strictly increasing and bounded above by 1, with supremum 1, but the value 1 is never attained since x = 1 is not in the domain; similarly, it has no minimum as values approach 0 without reaching it.^[39] Thus, despite continuity, f fails to achieve both its maximum and minimum on this non-compact domain. On an unbounded domain like the entire real line \mathbb{R}, continuous functions need not even be bounded. For instance, the identity function f(x) = x on \mathbb{R} is continuous but unbounded above (values go to +\infty) and below (values go to -\infty), so it attains neither a maximum nor a minimum.^[40] This illustrates how unboundedness prevents the boundedness required for extrema attainment. Even on a closed but unbounded interval, such as [0, \infty), a continuous function may be bounded yet fail to attain its infimum. The function f(x) = \frac{1}{1 + x^2} is continuous on [0, \infty), achieves its maximum of 1 at x = 0, but approaches an infimum of 0 as x \to \infty without ever reaching it, since f(x) > 0 for all x \geq 0.^[39] Here, the lack of compactness allows the function values to "escape" to the boundary at infinity, preventing full attainment of extrema.

Failure Due to Discontinuity

Discontinuity of a function on a compact domain can lead to the failure of either boundedness or the attainment of extrema, highlighting the necessity of the continuity assumption in the Extreme Value Theorem. While compact domains ensure certain topological properties, pathological behavior introduced by discontinuities can prevent the image from being bounded or from achieving its supremum and infimum within the domain. A standard example demonstrating failure of boundedness is the function defined on the compact interval [0,1] by

f(x) = \begin{cases} \frac{1}{x} & \text{if } 0 < x \leq 1, \\ 0 & \text{if } x = 0. \end{cases}

This function is discontinuous at x=0, as the right-hand limit \lim_{x \to 0^+} f(x) = +\infty does not equal f(0) = 0. The range of f is \{0\} \cup [1, +\infty), which is unbounded above, so no maximum exists.^[36] Even when a discontinuous function on [0,1] is bounded, it may fail to attain its extrema. Consider the function

g(x) = \begin{cases} x & \text{if } 0 \leq x < 1, \\ \frac{1}{2} & \text{if } x = 1. \end{cases}

Here, g is discontinuous at x=1, since \lim_{x \to 1^-} g(x) = 1 \neq \frac{1}{2} = g(1). The range is [0,1), which is bounded with infimum 0 (attained at x=0) but supremum 1 (not attained at any point in [0,1]).^[41] Another example where neither extremum is attained is

h(x) = \begin{cases} x & \text{if } 0.25 < x < 0.75, \\ 0.5 & \text{otherwise}. \end{cases}

This function is discontinuous at x=0.25 and x=0.75. The range is \{0.5\} \cup (0.25, 0.75), bounded with infimum 0.25 and supremum 0.75, neither of which is attained.^[42]

Proofs

Proof of Boundedness Theorem

To prove the Boundedness Theorem—that a continuous function f: K \to \mathbb{R} on a compact set K in \mathbb{R}^m is bounded—we proceed by contradiction, utilizing the definition of compactness via open covers and the subspace topology on K.^[43] Assume f is unbounded above on K. Consider the collection of sets \{V_n\}_{n=1}^\infty, where V_n = \{ x \in K \mid f(x) < n \}. Each V_n = f^{-1}((-\infty, n)) \cap K, and since (-\infty, n) is open in \mathbb{R} and f is continuous, f^{-1}((-\infty, n)) is open in \mathbb{R}^m, making V_n open in the relative (subspace) topology on K.^[43] This collection forms an open cover of K, as for any x \in K, f(x) is finite, so there exists n > f(x) with x \in V_n. By the assumption that f is unbounded above, for any finite subcollection \{V_{n_1}, \dots, V_{n_k}\}, let m = \max\{n_1, \dots, n_k\}. Then \bigcup_{i=1}^k V_{n_i} \subseteq V_m = \{ x \in K \mid f(x) < m \}. However, since f is unbounded above, there exists some x \in K with f(x) \geq m, so this x \notin V_m and thus not covered by the finite subcollection. Therefore, the open cover \{V_n\}_{n=1}^\infty has no finite subcover, contradicting the compactness of K.^[43] Hence, f must be bounded above. Similarly, to show bounded below, apply the same argument to -f, or consider the open cover \{ W_n \}_{n=1}^\infty where W_n = \{ x \in K \mid f(x) > -n \}; the assumption of unbounded below leads to an analogous contradiction. Thus, f is bounded on K.^[43] In the special case where K is a closed bounded interval [a, b] \subset \mathbb{R}, the Heine-Borel theorem establishes that such intervals are compact, providing the foundation for the argument above without needing the general subspace topology details.^[43]

Proof of Extreme Value Theorem

The Extreme Value Theorem asserts that if K is a compact subset of \mathbb{R}^n and f: K \to \mathbb{R} is continuous, then f attains its maximum and minimum values on K.^[8] To prove this, assume the Boundedness Theorem has established that the image f(K) is bounded, so let M = \sup f(K), which is finite.^[8] Construct a sequence \{x_n\} in K such that f(x_n) > M - 1/n for each n, ensuring f(x_n) \to M.^[6] Since K is compact, the Bolzano-Weierstrass theorem guarantees a subsequence \{x_{n_k}\} converging to some x \in K.^[8] By the continuity of f, it follows that f(x_{n_k}) \to f(x), and thus f(x) = M, so f attains its supremum at x.^[6] The argument for the infimum m = \inf f(K) proceeds analogously: select \{x_n\} with f(x_n) < m + 1/n, extract a convergent subsequence to x \in K, and apply continuity to obtain f(x) = m.^[8] Alternatively, since f is continuous and K is compact, the image f(K) is compact as the continuous image of a compact set.^[6] In \mathbb{R}, compact sets are closed and bounded, hence f(K) contains its supremum and infimum, confirming attainment.^[8] This proof relies solely on the axioms of \mathbb{R} and properties of compactness, without invoking derivatives or additional completeness assumptions beyond those for \mathbb{R}.^[6]

Non-Standard Proof Using Hyperreals

Nonstandard analysis offers an alternative proof of the Extreme Value Theorem by embedding the real numbers \mathbb{R} into the larger field of hyperreal numbers *\mathbb{R}, which includes infinitesimal and infinite quantities while preserving the ordered field structure via the transfer principle. This principle states that any first-order logical statement true in \mathbb{R} holds for *\mathbb{R} when interpreted in the nonstandard universe. For a compact set K \subseteq \mathbb{R}, the nonstandard extension *K satisfies the near-standard property: every h \in *K is infinitely close to some x \in K, meaning h \approx x or \text{st}(h) = x, where \text{st} denotes the standard part function mapping limited hyperreals to their unique real approximations. This characterization equates compactness with the condition *K \subseteq \bigcup_{x \in K} \text{hal}(x), where \text{hal}(x) is the monad of hyperreals infinitesimally close to x.^[44] Given a continuous function f: K \to \mathbb{R}, its nonstandard extension *f: *K \to *\mathbb{R} inherits continuity at near-standard points. The boundedness of f(K) transfers via the principle, implying *f(*K) is bounded in *\mathbb{R}. To attain the maximum, consider the supremum M = \sup \{ *f(h) \mid h \in *K \} in *\mathbb{R}; since K is compact, f is uniformly continuous, and there exists h \in *K such that *f(h) \approx M with h limited (finite). The transfer principle applies to the first-order statement that every nonempty finite subset of *\mathbb{R} has a maximum element, allowing a hyperfinite approximation of K (e.g., a hyperfinite grid covering K with infinitesimal mesh) where a maximum point h exists by internal set theory. Then, c = \text{st}(h) \in K, and uniform continuity ensures f(c) = \text{st}(*f(h)) = \sup f(K), so f(c) = \max f(K). For the specific case of K = [a, b], choose an infinite hypernatural N \in {}^*\mathbb{N} \setminus \mathbb{N} and the internal points p_k = a + k (b-a)/N for k = 0, \dots, N; transfer yields a k_0 maximizing *f(p_{k_0}), and d = \text{st}(p_{k_0}) gives the maximum of f.^[45]^[46] This nonstandard proof highlights the intuitive role of infinitesimals in capturing the "finest" behavior of continuous functions on compact sets, directly constructing maximizers via standard parts without explicit \epsilon-\delta arguments for limits or sequential compactness. It relies on the transfer principle to bridge standard and nonstandard realms, offering a conceptually simpler alternative to classical real analysis proofs while maintaining logical rigor through model-theoretic foundations.^[44]

Elementary Proof Using Nested Intervals and the Bisection Method

The elementary proof of the Extreme Value Theorem for a continuous function f: [a, b] \to \mathbb{R} relies on the least upper bound property of the real numbers to establish both boundedness and attainment of extrema, without invoking sequential compactness or open covers. Assuming the boundedness of f on [a, b] has been established separately (as in the proof of the Boundedness Theorem), let M = \sup \{ f(x) \mid x \in [a, b] \}, which exists by the least upper bound property since the image is nonempty and bounded above.^[47] To show that M is attained, construct a sequence of nested closed intervals I_n = [a_n, b_n] \subseteq [a, b] such that \sup f(I_n) = M for each n and the length b_n - a_n = (b - a)/2^n \to 0. Begin with I_1 = [a, b], where \sup f(I_1) = M. For each n, bisect I_n at the midpoint c_n = (a_n + b_n)/2 to form the left subinterval [a_n, c_n] and right subinterval [c_n, b_n]. Let M_L = \sup f([a_n, c_n]) and M_R = \sup f([c_n, b_n]), both of which exist by the least upper bound property applied to the bounded images f([a_n, c_n]) and f([c_n, b_n]). Since \max(M_L, M_R) = \sup f(I_n) = M, at least one subinterval has supremum M; select such a subinterval as I_{n+1}. This ensures the intervals are nested, closed, and contract in length to zero.^[47] By the nested interval property (equivalent to the least upper bound property for closed bounded intervals), the intersection \bigcap_{n=1}^\infty I_n is nonempty and consists of a single point c \in [a, b], where c = \sup \{ a_n \mid n \in \mathbb{N} \} = \inf \{ b_n \mid n \in \mathbb{N} \}. Since f is continuous at c, for any \epsilon > 0, there exists \delta > 0 such that if |x - c| < \delta, then |f(x) - f(c)| < \epsilon. Now, f(c) \leq M as M is an upper bound. Suppose for contradiction that f(c) < M; let \epsilon = (M - f(c))/2 > 0. Choose N large enough so that b_N - a_N < \delta. Then I_N \subseteq (c - \delta, c + \delta), implying f(x) < f(c) + \epsilon = (M + f(c))/2 < M for all x \in I_N. Thus, \sup f(I_N) < M, contradicting the construction. Therefore, f(c) = M. A symmetric argument applies to show the infimum is attained.^[47] This proof draws on the order-theoretic completeness of the reals via the least upper bound property and nested intervals, mirroring the bisection technique pioneered by Bernard Bolzano in his 1817 proof of the Intermediate Value Theorem, though adapted here for extrema without modern compactness notions.^[48] However, the approach is more verbose than sequential methods, requiring explicit interval construction at each step, and less readily generalizable to non-interval domains or non-ordered spaces, where topological compactness is needed instead.^[49]

Generalizations and Extensions

To Metric Spaces

The extreme value theorem generalizes to metric spaces in the following form: Let (X, d) be a metric space and K \subset X a compact subset. If f: K \to \mathbb{R} is continuous, then there exist points x_{\max}, x_{\min} \in K such that f(x_{\max}) = \max_{x \in K} f(x) and f(x_{\min}) = \min_{x \in K} f(x).^[50] This result holds because the image f(K) is compact in \mathbb{R}, and compact subsets of \mathbb{R} are closed and bounded, ensuring the supremum and infimum are attained.^[51] A proof sketch relies on the equivalence of compactness and sequential compactness in metric spaces, where every sequence in K has a convergent subsequence with limit in K.^[52] To establish boundedness, suppose f is unbounded above on K. Select a sequence \{x_n\} \subset K such that f(x_n) > n for each n. By sequential compactness, there exists a subsequence \{x_{n_k}\} converging to some x \in K. Continuity of f implies f(x_{n_k}) \to f(x), which is finite, contradicting f(x_{n_k}) \to \infty. A similar argument shows f is bounded below. For attainment of the maximum, let M = \sup_{x \in K} f(x). Choose \{x_n\} \subset K such that f(x_n) > M - 1/n. A convergent subsequence \{x_{n_k}\} \to x \in K yields f(x) = M by continuity. The minimum follows analogously.^[53] An illustrative example is the closed unit ball B = \{ x \in \mathbb{R}^n : \|x\| \leq 1 \} in the Euclidean metric, which is compact. Thus, any continuous function f: B \to \mathbb{R} attains its extrema on B. This compactness follows from the Heine-Borel theorem, which characterizes compact sets in \mathbb{R}^n as closed and bounded.^[54] In contrast to the real line, where compact sets are precisely the closed bounded intervals, the generalization to arbitrary metric spaces leverages sequential compactness without requiring an explicit embedding into \mathbb{R}. A key distinction is that continuity on compact metric spaces automatically implies uniform continuity, which strengthens the theorem's applicability in proofs involving approximations or extensions of functions.^[55] This property arises from the metric structure, allowing control over moduli of continuity via sequences and covers.^[56]

To Topological Spaces

The extreme value theorem extends naturally to topological spaces. Specifically, if X is a topological space, K \subset X is a compact subset, and f: K \to \mathbb{R} is continuous, then f(K) is compact in \mathbb{R}. Since compact subsets of \mathbb{R} are closed and bounded by the Heine-Borel theorem, f(K) is a closed bounded interval, and thus f attains both its maximum and minimum values on K.^[38]^[57] The key result underpinning this generalization is that the continuous image of a compact set is compact. This holds in any topological space without requiring a metric structure, relying instead on the open cover definition of compactness: if \{U_\alpha\} is an open cover of K, then continuity of f implies \{f^{-1}(U_\alpha)\} covers K, which has a finite subcover by compactness; the corresponding f(U_{\alpha_i}) then form a finite open cover of f(K). In Hausdorff spaces, such as \mathbb{R}, this ensures the image is closed, reinforcing the attainment of extrema.^[58]^[38] An illustrative example is the Cantor set C, a compact totally disconnected subspace of \mathbb{R}. As a compact topological space, any continuous function f: C \to \mathbb{R} must attain its extrema, demonstrating the theorem's applicability beyond connected domains like intervals, solely through topological compactness.^[38]

To Semi-Continuous Functions

A function f: X \to \mathbb{R}, where X is a topological space, is upper semi-continuous at a point a \in X if for every sequence \{x_n\} in X converging to a, \limsup_{n \to \infty} f(x_n) \leq f(a).^[59] Similarly, f is lower semi-continuous at a if \liminf_{n \to \infty} f(x_n) \geq f(a).^[59] These conditions weaken the requirement of full continuity, allowing functions with upward jumps for upper semi-continuity or downward jumps for lower semi-continuity, while still preserving certain extremal properties on suitable domains.^[59] If K is a nonempty compact subset of \mathbb{R}^n and f: K \to \mathbb{R} is upper semi-continuous, then f attains its maximum value on K, meaning there exists x^* \in K such that f(x^*) = \sup_{x \in K} f(x).^[59] Conversely, if f is lower semi-continuous on K, then f attains its minimum value on K, so there exists x^{**} \in K such that f(x^{**}) = \inf_{x \in K} f(x).^[59] Continuous functions satisfy both conditions simultaneously and thus attain both maxima and minima, aligning with the classical extreme value theorem.^[59] The proof for the upper semi-continuous case proceeds as follows: First, f is bounded above on the compact set K. To see this, suppose there exists a sequence \{x_k\} \subset K with f(x_k) \to +\infty; by compactness, a subsequence converges to some x \in K, but upper semi-continuity implies \limsup f(x_{k_j}) \leq f(x) < +\infty, a contradiction. Let s = \sup_{K} f < +\infty. The sets F_n = \{x \in K \mid f(x) \geq s - 1/n\} are nonempty for each n (as s is the supremum) and closed in K (since upper semi-continuity ensures that \{f < s - 1/n\} is open in K). These compact sets are nested, so their finite intersection property implies the intersection \bigcap_n F_n is nonempty; any point in this intersection satisfies f(x) = s. A symmetric argument applies to lower semi-continuous functions attaining minima, using the closedness of sublevel sets \{f \leq \alpha\}.^[60]^[61] Consider the function f: [0, 1] \to \mathbb{R} defined by f(0) = 1 and f(x) = x for x > 0. This function is upper semi-continuous on the compact interval [0, 1], as \limsup_{x \to 0} f(x) = 0 \leq f(0) = 1, permitting an upward jump at x = 0. The supremum is $1, attained at x = 0. However, f fails to be lower semi-continuous at x = 0 since \liminf_{x \to 0} f(x) = 0 < 1 = f(0), and indeed the infimum $0 is not attained. These results find applications in optimization problems involving inequalities, where objective functions may exhibit semi-continuity due to constraints or approximations; for instance, upper semi-continuous functions model maximization in distributionally robust optimization and variational analysis.

Historical Development

Early Contributions

In the late 17th century, Johann Bernoulli advanced the study of extrema through the calculus of variations by posing the brachistochrone problem in 1696, which sought the curve allowing the fastest descent under gravity between two points. This work implicitly assumed the existence of a continuous curve that attains a minimum time among all possible paths connecting the endpoints, without providing a proof of attainment. His brother Jacob Bernoulli solved the problem the following year using geometric methods inspired by Newton, again assuming that a minimizing extremum is attained for smooth, differentiable curves, though rigorous justification for continuity and boundedness was absent. Leonhard Euler built upon these foundations in the 18th century, particularly in his 1744 treatise Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes, where he formalized the calculus of variations and introduced the Euler-Lagrange equation as a necessary condition for extrema. Euler routinely assumed that continuous functions defined on finite intervals attain their maximum and minimum values, applying this intuition to problems in differential equations and variational principles, such as isoperimetric issues and geodesics on surfaces. This assumption facilitated practical solutions but relied on the geometric and infinitesimal approaches prevalent at the time, without addressing potential counterexamples for non-differentiable cases. The pre-analysis era's lack of rigor meant these contributions focused predominantly on differentiable functions, where extrema were sought via tangent conditions or infinitesimal variations, rather than general continuous functions. Proofs of existence were not pursued, as the emphasis was on computational methods for physical and geometric applications. These early ideas also intersected with precursors to the mean value theorem, notably Michel Rolle's 1691 theorem, which presupposed the attainment of an extremum between two points where a differentiable function takes equal values, laying groundwork for later analytic developments. Such assumptions foreshadowed the Bolzano-Weierstrass theorem in the following century.

Formalization in the 19th Century

In 1817, Bernard Bolzano employed the completeness property of the real numbers, specifically the least upper bound principle, in his proof of the intermediate value theorem for continuous functions on closed intervals. This approach implicitly established the boundedness of such functions, as Bolzano used the existence of suprema to construct sequences converging to roots, thereby demonstrating that continuous functions on bounded closed intervals cannot be unbounded. Although Bolzano's full treatment of the extreme value theorem appeared in his unpublished manuscript Functionenlehre from the 1830s,^[12] his 1817 work laid crucial groundwork by linking completeness to the attainment of bounds.^[62] During the 1860s, Karl Weierstrass advanced this foundation through his lectures at the University of Berlin, where he formalized the concepts of compactness and the extreme value theorem using what is now known as the Bolzano-Weierstrass theorem. In these lectures, Weierstrass proved that every bounded sequence in \mathbb{R} has a convergent subsequence, applying this to show that continuous functions on closed bounded intervals attain their maximum and minimum values. His epsilon-delta definition of continuity and emphasis on uniform convergence further solidified the rigorous framework for these results, disseminating them via student notes such as those from Hermann Amandus Schwarz in 1863/64.^[63]^[64] Augustin-Louis Cauchy contributed early ideas to uniform continuity in his 1821 Cours d'analyse, where he introduced precise notions of limits and continuity that influenced later developments, including covers of intervals through convergent sequences. Eduard Heine built on this in the 1870s, publishing the first explicit definition of uniform continuity in 1870 and proving in 1872 that continuous functions on closed bounded intervals are uniformly continuous, a key step toward understanding compactness via open covers.^[62]^[18] Weierstrass's ideas gained wider recognition in the 1880s through student publications and his own mimeographed lecture notes, such as the 1886 course on the foundations of function theory, which integrated the extreme value theorem into the axiomatic treatment of real analysis. These efforts established the theorem as a cornerstone of rigorous calculus.^[63] The 19th-century formalization of the extreme value theorem provided essential concepts of compactness and attainment of extrema that underpin modern topology, influencing later results like the Heine-Borel theorem.^[64]

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