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Continuous $f$ and integrable sign-changing $g$ with $\int_a^b fg \ne f(\xi)\int_a^b g$ for every $\xi$

Statement refuted

False claim: if $f$ is continuous on $[a,b]$ and $g$ is integrable on $[a,b]$ , then there is $\xi \in [a,b]$ with $\int_a^b f g = f(\xi)\int_a^b g$ .

That is If $f$ is continuous on $[a,b]$ and $g$ is integrable with $g \ge 0$ , there is $\xi \in [a,b]$ with $\int_a^b fg = f(\xi)\int_a^b g$ with the hypothesis $g \ge 0$ deleted, and it is false. On $[-1,1]$ take

$f(t) \;=\; t, \qquad g(t) \;=\; t .$

Both are continuous, hence integrable, and $g$ changes sign. Then

$\int_{-1}^{1} f g \;=\; \int_{-1}^{1} t^{2}\,\mathrm{d}t \;=\; \frac{2}{\iota(3)} \;>\; 0, \qquad \int_{-1}^{1} g \;=\; \int_{-1}^{1} t\,\mathrm{d}t \;=\; 0 ,$

so $f(\xi)\int_{-1}^{1} g = 0$ for every $\xi \in [-1,1]$ , while the left-hand side is positive. No $\xi$ works.

Facts & Assumptions

Given: The functions $f(t) = g(t) = t$ on $[-1,1]$ , and $\xi \in [-1,1]$ arbitrary.

[L1]

Every polynomial function is continuous, and a continuous function on $[-1,1]$ is integrable there (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$ , Darboux integrability as their equality, and the notation $\int_a^b f$ ).

[L2]

A product of two integrable functions on $[-1,1]$ is integrable (If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ , claim 1).

[L3]

For $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable at every real $c$ with derivative $\iota(n)c^{\,n-1}$ , and a scalar multiple of a differentiable function is differentiable with the scaled derivative (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ , claim 2, The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L4]

If $H$ is differentiable at every point of $[p,q]$ with $H'$ integrable there, then $\int_p^q H' = H(q)-H(p)$ ; a continuous function on an interval has a primitive (The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G' = f$ and $f$ is integrable, then $\int_a^b f = G(b)-G(a)$ , Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_a^b f = G(b)-G(a)$ for any primitive $G$ ).

[L5]

Powers and canonical naturals: $1^{2} = 1$ , $(-1)^{2} = 1$ , $1^{3} = 1$ , $(-1)^{3} = -1$ , $\iota(2) = 2 > 0$ and $\iota(3) = 3 > 0$ (Integer powers $a^m$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing, Ordered field).

[L6]

If $u \ge 0$ is integrable then $\int_p^q u \ge 0$ , and $u \cdot 0 = 0$ for every real $u$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ , Ordered field, Complete ordered field (least-upper-bound property), Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

Counterexample

technique · direct

1.1

$f$ and $g$ are continuous on $[-1,1]$ , hence integrable there by [L1], and $fg$ , the function $t \mapsto t^{2}$ , is integrable by [L1] or [L2].

givenL1L2

1.2

The function $H_1(t) := t^{3}/\iota(3)$ is differentiable at every real with $H_1'(t) = \iota(3)t^{2}/\iota(3) = t^{2}$ , by [L3] and [L5].

L3L5construct

1.3

The function $H_2(t) := t^{2}/\iota(2)$ is differentiable at every real with $H_2'(t) = \iota(2)\,t/\iota(2) = t$ , by [L3] and [L5].

L3L5construct

2.1

By [L4] applied to $H_1$ on $[-1,1]$ , $\int_{-1}^{1} t^{2}\,\mathrm{d}t = H_1(1)-H_1(-1) = 1/\iota(3) - (-1)/\iota(3) = 2/\iota(3)$ , a positive real by [L5].

step 1.1step 1.2L4L5

2.2

By [L4] applied to $H_2$ on $[-1,1]$ , $\int_{-1}^{1} t\,\mathrm{d}t = H_2(1)-H_2(-1) = 1/\iota(2) - 1/\iota(2) = 0$ .

step 1.1step 1.3L4L5

3.1

For every $\xi \in [-1,1]$ , $f(\xi)\int_{-1}^{1} g = \xi \cdot 0 = 0$ by step 2.2 and [L6], while $\int_{-1}^{1} fg = 2/\iota(3) > 0$ by step 2.1.

step 2.1step 2.2L5L6

4.1

Hence $\int_{-1}^{1} fg \ne f(\xi)\int_{-1}^{1}g$ for every $\xi \in [-1,1]$ , and the claim fails at this pair.

step 3.1L5∎

Remarks

What survives when the weight changes sign. The bound $\bigl|\int_a^b fg\bigr| \le \bigl(\sup_{[a,b]}|f|\bigr)\int_a^b|g|$ still holds, by If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ applied to $-\sup|f|\,|g| \le fg \le \sup|f|\,|g|$ together with If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ . What is lost is the identity: the weighted average $\int fg / \int g$ need not be a value of $f$ , and here it is not even defined.
Where the proof of the theorem breaks. With $g \ge 0$ the pointwise inequalities $mg \le fg \le Mg$ survive integration and bracket $\int_a^b fg$ between $m\int_a^b g$ and $M\int_a^b g$ (the pointwise inequalities are step 2.1 of If $f$ is continuous on $[a,b]$ and $g$ is integrable with $g \ge 0$ , there is $\xi \in [a,b]$ with $\int_a^b fg = f(\xi)\int_a^b g$ and the bracket is its step 3.1). Multiplying $m \le f \le M$ by a $g$ that changes sign reverses the inequality where $g < 0$ , so no such bracket is available, and the pair above shows that no weaker bracket can rescue the conclusion: the left-hand side is positive and the right-hand side is $0$ whatever $\xi$ is.
The theorem with $g \le 0$ holds too, by applying If $f$ is continuous on $[a,b]$ and $g$ is integrable with $g \ge 0$ , there is $\xi \in [a,b]$ with $\int_a^b fg = f(\xi)\int_a^b g$ to $-g$ and using linearity; what cannot be dropped is that $g$ has one sign.

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Sources

Mean value theorem (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)

Continuous fff and integrable sign-changing ggg with ∫abfg≠f(ξ)∫abg\int_a^b fg \ne f(\xi)\int_a^b g∫ab​fg=f(ξ)∫ab​g for every ξ\xiξ