Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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Bonnet's second mean value theorem: for f monotone and g integrable on [a,b] there is ξ∈[a,b] with ∫abfg=f(a)∫aξg+f(b)∫ξbg

Statement

Let a<b be reals, let f:[a,b]→R be monotone, that is nondecreasing or nonincreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences), and let g:[a,b]→R be integrable (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf). Then fg is integrable and there is ξ∈[a,b] with

∫abf g  =  f(a)∫aξg  +  f(b)∫ξbg.

No differentiability and no continuity of f is assumed. A monotone function may be discontinuous at infinitely many points and is still integrable (A monotone function on [a,b] is Riemann integrable: for the uniform partition into N parts the upper minus lower sum telescopes to ∣f(b)−f(a)∣ (b−a)/ι(N)), and the proof below uses only that its increments over the subintervals of a partition all have the same sign. This is the general form; the version usually proved by integration by parts needs f continuously differentiable, which is a strictly stronger hypothesis.

Facts & Assumptions

Given: Reals a<b, a monotone f:[a,b]→R, an integrable g:[a,b]→R, and a real ε>0. Write G(x):=∫axg for the integral function of g, and fix a real K≥0 with ∣g(t)∣≤K for every t∈[a,b].

[L2]

Products of integrable functions are integrable, as are absolute values, and ∣∫pqu∣≤∫pq∣u∣ for p≤q (If f,g are integrable on [a,b] then so are ∣f∣, f2, fg, max⁡(f,g) and min⁡(f,g), and ∣∫abf∣≤∫ab∣f∣).

[L5]

Abel summation by parts: with Aj=∑k<jαk, for every n≥1 one has ∑k<nαkβk=Anβn−1−∑k<n−1Ak+1(βk+1−βk) (Abel summation by parts: with An=∑k<nak one has ∑k<nakbk=Anbn−1−∑k<n−1Ak+1 (bk+1−bk) for every n≥1, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

Finite sums: additivity, scaling, splitting with the shift ∑k=pq−1xk=∑j<q−pxp+j, monotonicity in the terms, and telescoping ∑k<n(ck+1−ck)=cn−c0 (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

For a partition P=(n,t) of [a,b]: n≥1, t0=a, tk=b for k≥n, Δi=ti+1−ti>0 for i<n, Ii=[ti,ti+1]⊆[a,b], and U(f,P)−L(f,P)=∑i<n(Mi−mi)Δi with mi≤f(x)≤Mi for x∈Ii (Partition of [a,b] as a finite strictly increasing list a=t0<t1<⋯<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, For bounded f on [a,b] and a partition P: the infimum mi and supremum Mi of f on the i-th subinterval, and the lower and upper Darboux sums L(f,P)=∑imiΔi and U(f,P)=∑iMiΔi).

[L8]

Riemann's criterion for the integrable f: for every real η>0 there is a partition P with U(f,P)−L(f,P)<η (Riemann's criterion: a bounded f on [a,b] is Darboux integrable if and only if for every real ε>0 there is a partition P with U(f,P)−L(f,P)<ε).

[L10]

Absolute value and ordered-field arithmetic: −c≤x≤c is equivalent to ∣x∣≤c, multiplying an inequality by a positive real preserves it and by a negative real reverses it, the order is total and transitive, and a real that is ≤M+η for every real η>0 is ≤M (Basic properties of the absolute value, Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct
1.1

f is bounded and integrable by [L1], so fg is integrable by [L2]; put Ψ(x):=∫axfg, so Ψ(b)=∫abfg and Ψ(y)−Ψ(x)=∫xyfg by [L3] applied to fg.

givenL1L2L3
1.2

G is continuous on [a,b], so by [L4] there are m≤M with G[ [a,b] ]=[m,M], m=min⁡G[ [a,b] ] and M=max⁡G[ [a,b] ].

L3L4choose
1.3

Put C:=f(b)−f(a) and, for a partition P=(n,t) of [a,b], put dj:=f(tj+1)−f(tj) for j<n. By [L6], ∑j<ndj=f(tn)−f(t0)=C; and all the dj are ≥0 when f is nondecreasing and all are ≤0 when f is nonincreasing, by [L7] and Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences.

givenL6L7construct
2.1

The case C=0. Then f(a)=f(b), and monotonicity forces f(x)=f(a) for every x∈[a,b], since f(x) lies between f(a) and f(b); so ∫abfg=f(a)∫abg by [L9], while the right-hand side at ξ:=a is f(a)⋅0+f(b)∫abg=f(a)∫abg by [L3]. The theorem holds with ξ=a.

step 1.1step 1.3L3L9L10
2.2

Abel summation on a partition. Let P=(n,t) be any partition of [a,b]. Apply [L5] with αk:=G(tk+1)−G(tk) and βk:=f(tk+1), noting tk∈[a,b] for every k∈N by [L7]. By [L6], Aj=∑k<j(G(tk+1)−G(tk))=G(tj)−G(t0)=G(tj), since G(a)=0.

step 1.2L3L5L6L7
2.3

S(P) approximates ∫abfg. For k<n one has G(tk+1)−G(tk)=∫tktk+1g by [L3], so by [L9] the k-th term of Ψ(tk+1)−Ψ(tk)−f(tk+1)(G(tk+1)−G(tk)) equals ∫tktk+1(f−f(tk+1))g.

step 1.1L3L9
3.1

So, writing S(P):=∑k<n(G(tk+1)−G(tk))f(tk+1), [L5] gives S(P)=G(tn)f(tn)−∑k<n−1G(tk+1)(f(tk+2)−f(tk+1))=G(b)f(b)−∑k<n−1xk+1, where xj:=G(tj) dj.

step 2.2L5L7construct
3.2

For x∈Ik both f(x) and f(tk+1) lie in [mk,Mk], so ∣(f(x)−f(tk+1))g(x)∣≤K (Mk−mk); hence by [L2] and [L9], −K(Mk−mk)Δk≤∫tktk+1(f−f(tk+1))g≤K(Mk−mk)Δk.

step 2.3givenL2L7L9L10
4.1

By [L6], ∑j<nxj=x0+∑k<n−1xk+1, and x0=G(t0)d0=G(a)d0=0; so, putting T(P):=∑j<nG(tj) dj, step 3.1 reads S(P)=G(b)f(b)−T(P).

step 3.1L3L6construct
4.2

Summing over k<n with [L6], and telescoping ∑k<n(Ψ(tk+1)−Ψ(tk))=Ψ(b)−Ψ(a)=∫abfg, gives ∣∫abfg−S(P)∣≤K(U(f,P)−L(f,P)).

step 1.1step 2.3step 3.2L6L7L10
5.1

T(P) is λPC for some λP∈[m,M], when C≠0. By step 1.2, m≤G(tj)≤M for every j. If f is nondecreasing then dj≥0, so m dj≤G(tj)dj≤M dj, and summing with [L6] and step 1.3 gives mC≤T(P)≤MC with C≥0; if f is nonincreasing then dj≤0, so M dj≤G(tj)dj≤m dj, and summing gives MC≤T(P)≤mC with C≤0. Dividing by C in the first case, and by the negative C with the inequalities reversed in the second, gives m≤T(P)/C≤M in both.

step 1.2step 1.3step 4.1L6L10construct
6.1

The case C≠0. Put λ:=(G(b)f(b)−∫abfg)/C. By step 4.1, ∫abfg−S(P)=∫abfg−G(b)f(b)+T(P)=(λP−λ)C for every partition P, where λP=T(P)/C.

step 4.1step 5.1L10construct
7.1

By [L8] fix a partition P with U(f,P)−L(f,P)<ε ∣C∣/(K+1), a positive real; then step 4.2 and step 6.1 give ∣λP−λ∣ ∣C∣≤K(U(f,P)−L(f,P))<ε∣C∣, so ∣λP−λ∣<ε.

step 4.2step 6.1L8L10choose
8.1

Since m≤λP≤M by step 5.1, it follows that m−ε<λ<M+ε; as ε>0 was arbitrary, m≤λ≤M.

step 5.1step 7.1L10
9.1

By step 1.2, G[ [a,b] ]=[m,M], so there is ξ∈[a,b] with G(ξ)=λ.

step 1.2step 8.1L4choose
10.1

Then ∫abfg=G(b)f(b)−λC=f(b)G(b)−G(ξ)(f(b)−f(a))=f(a)G(ξ)+f(b)(G(b)−G(ξ)), and G(ξ)=∫aξg with G(b)−G(ξ)=∫ξbg by [L3]; this is the stated identity.

step 6.1step 9.1L3algebra
11.1

The cases C=0 and C≠0 are exhaustive, so the theorem holds in both.

step 2.1step 10.1L10∎

Remarks

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