{"history":[],"article":{"id":"c00957d3-e55d-4674-896b-a6012ae9fe08","title":"Mean value theorem","slug":"mean-value-theorem","content":":::summary\n:::\n\n# \nIn [mathematics](https://bliptext.com/articles/mathematics), the **mean value theorem** (or Lagrange's **mean value theorem**) states, roughly, that for a given planar [arc](https://bliptext.com/articles/arc-geometry) between two endpoints, there is at least one point at which the [tangent](https://bliptext.com/articles/tangent) to the [arc](https://bliptext.com/articles/arc-geometry) is parallel to the [secant](https://bliptext.com/articles/secant-line) through its endpoints.\nIt is one of the most important results in [real analysis](https://bliptext.com/articles/real-analysis).\nThis theorem is used to prove statements about a function on an [interval](https://bliptext.com/articles/interval-mathematics) starting from local hypotheses about derivatives at points of the [interval](https://bliptext.com/articles/interval-mathematics).\n\n# History\nA special case of this [theorem](https://bliptext.com/articles/theorem) for inverse interpolation of the sine was first described by [Parameshvara](https://bliptext.com/articles/parameshvara) (1380–1460), from the [Kerala School of Astronomy and Mathematics](https://bliptext.com/articles/kerala-school-of-astronomy-and-mathematics) in [India](https://bliptext.com/articles/india), in his commentaries on [Govindasvāmi](https://bliptext.com/articles/govindasv-mi) and [Bhāskara II](https://bliptext.com/articles/bh-skara-ii).\nA restricted form of the theorem was proved by [Michel Rolle](https://bliptext.com/articles/michel-rolle) in 1691; the result was what is now known as [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem), and was proved only for polynomials, without the techniques of calculus.\nThe mean value theorem in its modern form was stated and proved by [Augustin Louis Cauchy](https://bliptext.com/articles/augustin-louis-cauchy) in 1823.\nMany variations of this theorem have been proved since then.\n\n# Statement\nLet be a [continuous function](https://bliptext.com/articles/continuous-function) on the [closed interval](https://bliptext.com/articles/closed-interval) [a,b], and [differentiable](https://bliptext.com/articles/differentiable-function) on the [open interval](https://bliptext.com/articles/open-interval) (a,b), where a < b. Then there exists some c in (a,b) such that:\nThe mean value theorem is a generalization of [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem), which assumes f(a)=f(b), so that the right-hand side above is zero.\nThe mean value theorem is still valid in a slightly more general setting.\nOne only needs to assume that is [continuous](https://bliptext.com/articles/continuous-function) on [a,b], and that for every x in (a,b) the [limit](https://bliptext.com/articles/limit-of-a-function)\nexists as a finite number or equals \\infty or -\\infty.\nIf finite, that limit equals f'(x).\nAn example where this version of the theorem applies is given by the real-valued [cube root](https://bliptext.com/articles/cube-root) function mapping, whose [derivative](https://bliptext.com/articles/derivative) tends to infinity at the origin.\n# [The function f attains the slope of the secant between a and b as the derivative at the point \\xi\\in(a,b). | upload.wikimedia.org/wikipedia/commons/9/94/Mittelwertsatz3.svg]\n# [It is also possible that there are multiple tangents parallel to the secant. | upload.wikimedia.org/wikipedia/commons/f/f9/Mittelwertsatz6.svg]\n\n## Proof\nThe expression gives the [slope](https://bliptext.com/articles/slope) of the line joining the points (a,f(a)) and (b,f(b)), which is a [chord](https://bliptext.com/articles/chord-geometry) of the graph of f, while f'(x) gives the [slope](https://bliptext.com/articles/slope) of the tangent to the curve at the point (x,f(x)).\nThus the mean value theorem says that given any chord of a smooth curve, we can find a point on the curve lying between the end-points of the chord such that the tangent of the curve at that point is parallel to the chord.\nThe following proof illustrates this idea.\nDefine, where r is a constant.\nSince f is continuous on [a,b] and differentiable on (a,b), the same is true for g. We now want to choose r so that g satisfies the conditions of [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem).\nNamely\nBy [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem), since g is differentiable and g(a)=g(b), there is some c in (a,b) for which g'(c)=0, and it follows from the equality that,\n\n# Implications\n**Theorem 1:** Assume that f is a continuous, real-valued function, defined on an arbitrary interval I of the real line.\nIf the derivative of f at every [interior point](https://bliptext.com/articles/interior-topology) of the interval I exists and is zero, then f is [constant](https://bliptext.com/articles/constant-function) in the interior.\n**Proof:** Assume the derivative of f at every [interior point](https://bliptext.com/articles/interior-topology) of the interval I exists and is zero.\nLet (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a, b) such that\nThis implies that f(a) = f(b).\nThus, f is constant on the interior of I and thus is constant on I by continuity.\n(See below for a multivariable version of this result.)\n**Remarks:**\n**Theorem 2:** If for all x in an interval (a, b) of the domain of these functions, then f - g is constant, i.e. f = g + c where c is a constant on (a, b).\n**Proof:** Let, then on the interval (a, b), so the above theorem 1 tells that is a constant c or f = g + c.\n**Theorem 3:** If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + c where c is a constant.\n**Proof:** It directly follows from the theorem 2 above.\n\n# Cauchy's mean value theorem\n**Cauchy's mean value theorem**, also known as the **extended mean value theorem**, is a generalization of the mean value theorem.\nIt states: if the functions f and g are both continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists some c \\in (a,b), such that\nOf course, if and, this is equivalent to:\nGeometrically, this means that there is some [tangent](https://bliptext.com/articles/tangent) to the graph of the [curve](https://bliptext.com/articles/curve)\nwhich is [parallel](https://bliptext.com/articles/parallel-geometry) to the line defined by the points and.\nHowever, Cauchy's theorem does not claim the existence of such a tangent in all cases where and are distinct points, since it might be satisfied only for some value c with, in other words a value for which the mentioned curve is [stationary](https://bliptext.com/articles/stationary-point); in such points no tangent to the curve is likely to be defined at all.\nAn example of this situation is the curve given by\nwhich on the interval [-1,1] goes from the point (-1, 0) to (1, 0), yet never has a horizontal tangent; however it has a stationary point (in fact a [cusp](https://bliptext.com/articles/cusp-singularity)) at t = 0.\nCauchy's mean value theorem can be used to prove [L'Hôpital's rule](https://bliptext.com/articles/l-h-pital-s-rule).\nThe mean value theorem is the special case of Cauchy's mean value theorem when g(t) = t.\n# [Geometrical meaning of Cauchy's theorem | upload.wikimedia.org/wikipedia/commons/7/7a/Cauchy.svg]\n\n## Proof\nThe proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.\n

\nSuppose.\nDefine, where r is fixed in such a way that h(a) = h(b), namely\nSince f and g are continuous on [a,b] and differentiable on (a,b), the same is true for h. All in all, h satisfies the conditions of [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem): consequently, there is some c in (a,b) for which h'(c) = 0.\nNow using the definition of h we have:\nand thus\n
\nIf g(a) = g(b), then, applying [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem) to g, it follows that there exists c in (a,b) for which g'(c) = 0.\nUsing this choice of c, Cauchy's mean value theorem (trivially) holds.\n

\n\n# Mean value theorem in several variables\nThe mean value theorem generalizes to real functions of multiple variables.\nThe trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.\nLet G be an open subset of \\R^n, and let f:G\\to\\R be a differentiable function.\nFix points x,y\\in G such that the line segment between x, y lies in G, and define.\nSince g is a differentiable function in one variable, the mean value theorem gives:\nfor some c between 0 and 1.\nBut since g(1)=f(y) and g(0)=f(x), computing g'(c) explicitly we have:\nwhere \\nabla denotes a [gradient](https://bliptext.com/articles/gradient) and \\cdot a [dot product](https://bliptext.com/articles/dot-product).\nThis is an exact analog of the theorem in one variable (in the case n=1 this is the theorem in one variable).\nBy the [Cauchy–Schwarz inequality](https://bliptext.com/articles/cauchy-schwarz-inequality), the equation gives the estimate:\nIn particular, when G is convex and the partial derivatives of f are bounded, f is [Lipschitz continuous](https://bliptext.com/articles/lipschitz-continuity) (and therefore [uniformly continuous](https://bliptext.com/articles/uniform-continuity)).\nAs an application of the above, we prove that f is constant if the open subset G is connected and every partial derivative of f is 0.\nPick some point x_0\\in G, and let.\nWe want to show g(x)=0 for every x\\in G. For that, let.\nThen E is closed in G and nonempty.\nIt is open too: for every x\\in E ,\nfor every y in open ball centered at x and contained in G. Since G is connected, we conclude E=G.\nThe above arguments are made in a coordinate-free manner; hence, they generalize to the case when G is a subset of a Banach space.\n\n# Mean value theorem for vector-valued functions\nThere is no exact analog of the mean value theorem for vector-valued functions (see below).\nHowever, there is an inequality which can be applied to many of the same situations to which the mean value theorem is applicable in the one dimensional case:\n\n## Mean value inequality\n[Jean Dieudonné](https://bliptext.com/articles/jean-dieudonn) in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and one cannot find the mean value and in applications one only needs mean inequality.\n[Serge Lang](https://bliptext.com/articles/serge-lang) in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative.\nIf one uses the [Henstock–Kurzweil integral](https://bliptext.com/articles/henstock-kurzweil-integral) one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock–Kurzweil integrable.\nThe reason why there is no analog of mean value equality is the following: If\nf : U → Rm\nis a differentiable function (where\nU ⊂ Rn\nis open) and if\nx + th\n,\nx, h ∈ Rn, t ∈ [0, 1]\nis the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions\nfi (i = 1, …, m)\nof f (in the above notation set\ny = x + h\n). In doing so one finds points\nx + tih\non the line segment satisfying\nBut generally there will not be a single point\nx + t*h\non the line segment satisfying\nfor all i simultaneously.\nFor example, define:\nThen, but and are never simultaneously zero as x ranges over.\nThe above theorem implies the following:\nIn fact, the above statement suffices for many applications and can be proved directly as follows.\n(We shall write f for \\textbf{f} for readability.)\n\n# Cases where the theorem cannot be applied\nAll conditions for the mean value theorem are necessary:\nWhen one of the above conditions is not satisfied, the mean value theorem is not valid in general, and so it cannot be applied.\nThe necessity of the first condition can be seen by the counterexample where the function f(x)=|x| on [-1,1] is not differentiable.\nThe necessity of the second condition can be seen by the counterexample where the function\nsatisfies criteria 1 since f'(x)=0 on (0,1) but not criteria 2 since\nand for all x\\in (0,1) so no such c exists.\nThe theorem is false if a differentiable function is complex-valued instead of real-valued.\nFor example, if for all real x, then\nwhile f'(x)\\ne 0 for any real x.\n# [mean value theorem | upload.wikimedia.org/wikipedia/commons/0/07/Absolute///Value.svg]\n\n# Mean value theorems for definite integrals\n\n## First mean value theorem for definite integrals\nLet f : [a, b] → **R** be a continuous function.\nThen there exists c in (a, b) such that\nThis follows at once from the [fundamental theorem of calculus](https://bliptext.com/articles/fundamental-theorem-of-calculus), together with the mean value theorem for derivatives.\nSince the mean value of f on [a, b] is defined as\nwe can interpret the conclusion as f achieves its mean value at some c in (a, b).\nIn general, if f : [a, b] → **R** is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that\n# [Geometrically: interpreting f(c) as the height of a rectangle and b–a as the width, this rectangle has the same area as the region below the curve from a to b | upload.wikimedia.org/wikipedia/commons/5/5b/%E7%A7%AF%E5%88%86%E4%B8%AD%E5%80%BC%E5%AE%9A%E7%90%86.jpg]\n\n## Second mean value theorem for definite integrals\nThere are various slightly different theorems called the **second mean value theorem for definite integrals**.\nA commonly found version is as follows:\nHere G(a^+) stands for, the existence of which follows from the conditions.\nNote that it is essential that the interval (a, b] contains b. A variant not having this requirement is:\nIf the function G returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of G is also multi-dimensional.\nFor example, consider the following 2-dimensional function defined on an n-dimensional cube:\nThen, by symmetry it is easy to see that the mean value of G over its domain is (0,0):\nHowever, there is no point in which G=(0,0), because |G|=1 everywhere.\n\n# Generalizations\n\n## Linear algebra\nAssume that f, g, and h are differentiable functions on (a,b) that are continuous on [a,b].\nDefine\nThere exists c\\in(a,b) such that D'(c)=0.\nNotice that\nand if we place h(x)=1, we get **Cauchy's mean value theorem**.\nIf we place h(x)=1 and g(x)=x we get **Lagrange's mean value theorem**.\nThe proof of the generalization is quite simple: each of D(a) and D(b) are [determinants](https://bliptext.com/articles/determinant) with two identical rows, hence D(a)=D(b)=0.\nThe Rolle's theorem implies that there exists c\\in (a,b) such that D'(c)=0.\n\n## Probability theory\nLet X and Y be non-negative [random variables](https://bliptext.com/articles/random-variable) such that E[X] < E[Y] < ∞ and (i.e. X is smaller than Y in the [usual stochastic order](https://bliptext.com/articles/stochastic-ordering)).\nThen there exists an absolutely continuous non-negative random variable Z having [probability density function](https://bliptext.com/articles/probability-density-function)\nLet g be a [measurable](https://bliptext.com/articles/measurable-function) and [differentiable function](https://bliptext.com/articles/differentiable-function) such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be [measurable](https://bliptext.com/articles/measurable-function) and [Riemann-integrable](https://bliptext.com/articles/riemann-integral) on the interval [x, y] for all y ≥ x ≥ 0.\nThen, E[g′(Z)] is finite and\n\n## Complex analysis\nAs noted above, the theorem does not hold for differentiable complex-valued functions.\nInstead, a generalization of the theorem is stated such:\nLet f : Ω → **C** be a [holomorphic function](https://bliptext.com/articles/holomorphic-function) on the open convex set Ω, and let a and b be distinct points in Ω.\nThen there exist points u, v on the interior of the line segment from a to b such that\nWhere Re is the real part and Im is the imaginary part of a complex-valued function.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T17:26:46.358Z","updated_at":"2025-01-25T17:26:46.358Z","search_vector":"'mean':1 'theorem':3 'valu':2","search_content":" \n \n\n# \nIn [mathematics](https://bliptext.com/articles/mathematics), the **mean value theorem** (or Lagrange's **mean value theorem**) states, roughly, that for a given planar [arc](https://bliptext.com/articles/arc-geometry) between two endpoints, there is at least one point at which the [tangent](https://bliptext.com/articles/tangent) to the [arc](https://bliptext.com/articles/arc-geometry) is parallel to the [secant](https://bliptext.com/articles/secant-line) through its endpoints.\nIt is one of the most important results in [real analysis](https://bliptext.com/articles/real-analysis).\nThis theorem is used to prove statements about a function on an [interval](https://bliptext.com/articles/interval-mathematics) starting from local hypotheses about derivatives at points of the [interval](https://bliptext.com/articles/interval-mathematics).\n\n# History\nA special case of this [theorem](https://bliptext.com/articles/theorem) for inverse interpolation of the sine was first described by [Parameshvara](https://bliptext.com/articles/parameshvara) (1380–1460), from the [Kerala School of Astronomy and Mathematics](https://bliptext.com/articles/kerala-school-of-astronomy-and-mathematics) in [India](https://bliptext.com/articles/india), in his commentaries on [Govindasvāmi](https://bliptext.com/articles/govindasv-mi) and [Bhāskara II](https://bliptext.com/articles/bh-skara-ii).\nA restricted form of the theorem was proved by [Michel Rolle](https://bliptext.com/articles/michel-rolle) in 1691; the result was what is now known as [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem), and was proved only for polynomials, without the techniques of calculus.\nThe mean value theorem in its modern form was stated and proved by [Augustin Louis Cauchy](https://bliptext.com/articles/augustin-louis-cauchy) in 1823.\nMany variations of this theorem have been proved since then.\n\n# Statement\nLet be a [continuous function](https://bliptext.com/articles/continuous-function) on the [closed interval](https://bliptext.com/articles/closed-interval) [a,b], and [differentiable](https://bliptext.com/articles/differentiable-function) on the [open interval](https://bliptext.com/articles/open-interval) (a,b), where a \n \nSuppose.\nDefine, where r is fixed in such a way that h(a) = h(b), namely\nSince f and g are continuous on [a,b] and differentiable on (a,b), the same is true for h. All in all, h satisfies the conditions of [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem): consequently, there is some c in (a,b) for which h'(c) = 0.\nNow using the definition of h we have:\nand thus\n \n \nIf g(a) = g(b), then, applying [Rolle's theorem](https://bliptext.com/articles/rolle-s-theorem) to g, it follows that there exists c in (a,b) for which g'(c) = 0.\nUsing this choice of c, Cauchy's mean value theorem (trivially) holds.\n \n \n\n# Mean value theorem in several variables\nThe mean value theorem generalizes to real functions of multiple variables.\nThe trick is to use parametrization to create a real function of one variable, and then apply the one-variable theorem.\nLet G be an open subset of \\R^n, and let f:G\\to\\R be a differentiable function.\nFix points x,y\\in G such that the line segment between x, y lies in G, and define.\nSince g is a differentiable function in one variable, the mean value theorem gives:\nfor some c between 0 and 1.\nBut since g(1)=f(y) and g(0)=f(x), computing g'(c) explicitly we have:\nwhere \\nabla denotes a [gradient](https://bliptext.com/articles/gradient) and \\cdot a [dot product](https://bliptext.com/articles/dot-product).\nThis is an exact analog of the theorem in one variable (in the case n=1 this is the theorem in one variable).\nBy the [Cauchy–Schwarz inequality](https://bliptext.com/articles/cauchy-schwarz-inequality), the equation gives the estimate:\nIn particular, when G is convex and the partial derivatives of f are bounded, f is [Lipschitz continuous](https://bliptext.com/articles/lipschitz-continuity) (and therefore [uniformly continuous](https://bliptext.com/articles/uniform-continuity)).\nAs an application of the above, we prove that f is constant if the open subset G is connected and every partial derivative of f is 0.\nPick some point x_0\\in G, and let.\nWe want to show g(x)=0 for every x\\in G. For that, let.\nThen E is closed in G and nonempty.\nIt is open too: for every x\\in E ,\nfor every y in open ball centered at x and contained in G. Since G is connected, we conclude E=G.\nThe above arguments are made in a coordinate-free manner; hence, they generalize to the case when G is a subset of a Banach space.\n\n# Mean value theorem for vector-valued functions\nThere is no exact analog of the mean value theorem for vector-valued functions (see below).\nHowever, there is an inequality which can be applied to many of the same situations to which the mean value theorem is applicable in the one dimensional case:\n\n## Mean value inequality\n[Jean Dieudonné](https://bliptext.com/articles/jean-dieudonn) in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and one cannot find the mean value and in applications one only needs mean inequality.\n[Serge Lang](https://bliptext.com/articles/serge-lang) in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative.\nIf one uses the [Henstock–Kurzweil integral](https://bliptext.com/articles/henstock-kurzweil-integral) one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock–Kurzweil integrable.\nThe reason why there is no analog of mean value equality is the following: If\nf : U → Rm\nis a differentiable function (where\nU ⊂ Rn\nis open) and if\nx + th\n,\nx, h ∈ Rn, t ∈ [0, 1]\nis the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions\nfi (i = 1, …, m)\nof f (in the above notation set\ny = x + h\n). In doing so one finds points\nx + tih\non the line segment satisfying\nBut generally there will not be a single point\nx + t*h\non the line segment satisfying\nfor all i simultaneously.\nFor example, define:\nThen, but and are never simultaneously zero as x ranges over.\nThe above theorem implies the following:\nIn fact, the above statement suffices for many applications and can be proved directly as follows.\n(We shall write f for \\textbf{f} for readability.)\n\n# Cases where the theorem cannot be applied\nAll conditions for the mean value theorem are necessary:\nWhen one of the above conditions is not satisfied, the mean value theorem is not valid in general, and so it cannot be applied.\nThe necessity of the first condition can be seen by the counterexample where the function f(x)=|x| on [-1,1] is not differentiable.\nThe necessity of the second condition can be seen by the counterexample where the function\nsatisfies criteria 1 since f'(x)=0 on (0,1) but not criteria 2 since\nand for all x\\in (0,1) so no such c exists.\nThe theorem is false if a differentiable function is complex-valued instead of real-valued.\nFor example, if for all real x, then\nwhile f'(x)\\ne 0 for any real x.\n# [mean value theorem | upload.wikimedia.org/wikipedia/commons/0/07/Absolute///Value.svg]\n\n# Mean value theorems for definite integrals\n\n## First mean value theorem for definite integrals\nLet f : [a, b] → **R** be a continuous function.\nThen there exists c in (a, b) such that\nThis follows at once from the [fundamental theorem of calculus](https://bliptext.com/articles/fundamental-theorem-of-calculus), together with the mean value theorem for derivatives.\nSince the mean value of f on [a, b] is defined as\nwe can interpret the conclusion as f achieves its mean value at some c in (a, b).\nIn general, if f : [a, b] → **R** is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that\n# [Geometrically: interpreting f(c) as the height of a rectangle and b–a as the width, this rectangle has the same area as the region below the curve from a to b | upload.wikimedia.org/wikipedia/commons/5/5b/%E7%A7%AF%E5%88%86%E4%B8%AD%E5%80%BC%E5%AE%9A%E7%90%86.jpg]\n\n## Second mean value theorem for definite integrals\nThere are various slightly different theorems called the **second mean value theorem for definite integrals**.\nA commonly found version is as follows:\nHere G(a^+) stands for, the existence of which follows from the conditions.\nNote that it is essential that the interval (a, b] contains b. A variant not having this requirement is:\nIf the function G returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of G is also multi-dimensional.\nFor example, consider the following 2-dimensional function defined on an n-dimensional cube:\nThen, by symmetry it is easy to see that the mean value of G over its domain is (0,0):\nHowever, there is no point in which G=(0,0), because |G|=1 everywhere.\n\n# Generalizations\n\n## Linear algebra\nAssume that f, g, and h are differentiable functions on (a,b) that are continuous on [a,b].\nDefine\nThere exists c\\in(a,b) such that D'(c)=0.\nNotice that\nand if we place h(x)=1, we get **Cauchy's mean value theorem**.\nIf we place h(x)=1 and g(x)=x we get **Lagrange's mean value theorem**.\nThe proof of the generalization is quite simple: each of D(a) and D(b) are [determinants](https://bliptext.com/articles/determinant) with two identical rows, hence D(a)=D(b)=0.\nThe Rolle's theorem implies that there exists c\\in (a,b) such that D'(c)=0.\n\n## Probability theory\nLet X and Y be non-negative [random variables](https://bliptext.com/articles/random-variable) such that E[X] < E[Y] < ∞ and (i.e. X is smaller than Y in the [usual stochastic order](https://bliptext.com/articles/stochastic-ordering)).\nThen there exists an absolutely continuous non-negative random variable Z having [probability density function](https://bliptext.com/articles/probability-density-function)\nLet g be a [measurable](https://bliptext.com/articles/measurable-function) and [differentiable function](https://bliptext.com/articles/differentiable-function) such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be [measurable](https://bliptext.com/articles/measurable-function) and [Riemann-integrable](https://bliptext.com/articles/riemann-integral) on the interval [x, y] for all y ≥ x ≥ 0.\nThen, E[g′(Z)] is finite and\n\n## Complex analysis\nAs noted above, the theorem does not hold for differentiable complex-valued functions.\nInstead, a generalization of the theorem is stated such:\nLet f : Ω → **C** be a [holomorphic function](https://bliptext.com/articles/holomorphic-function) on the open convex set Ω, and let a and b be distinct points in Ω.\nThen there exist points u, v on the interior of the line segment from a to b such that\nWhere Re is the real part and Im is the imaginary part of a complex-valued function.","revisionCount":0},"hasMore":false,"nextCursor":null}