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This is an attempt to generalize

Inspired by that question and the given answers, I have the following

Conjecture: Let $c:[a, b] \to \Bbb R$ be a continuous function. Then $$ \tag{1} \int_a^b f(x) c(x) \, dx \ge 0 $$ holds for all convex functions $f:[a, b] \to \Bbb R$ if and only if $$ \tag{2} \int_a^b c(x) \,dx = \int_a^b x c(x) \, dx = 0 \, . $$

The “only if” direction is easy: If $(1)$ holds for the four convex functions $x \mapsto \pm 1$, $x \mapsto \pm x$ then $(2)$ holds.

So the interesting part is the “if” direction. In the above mentioned Q&A this has been proven for the functions $c(x) = \cos(k x)$ (on the interval $[0, 2 \pi]$). Some of the proofs given there use the fact that $$ \int_0^{2\pi} \cos (x) \, dx = \int_0^{2\pi} x \cos (x) \, dx = 0 \, , $$ but all proofs use also symmetries of the cosine function, trigonometric identities, or where the cosine is positive and negative in $[0, 2 \pi]$. My conjecture is that these additional properties of the cosine are not needed, and that $(2)$ alone is sufficient to prove $(1)$ for all convex functions $f$.

As pointed out in the comments, the following is wrong: There is a simple proof for the “if” direction under the additional assumption that $f$ is twice continuously differentiable: Let $c_1$ be an antiderivative of $c$, and $c_2$ be an antiderivative of $c_1$. By adding a constant to $c_2$, if necessary, we can assume that $c_2(x) \ge 0$ on $[a, b]$. If $(1)$ holds then $c_1(a) = c_1(b)$ and $c_2(a) = c_2(b)$, and for all convex functions $f$ on $[a, b]$ is, using integration by parts (twice): $$ \int_a^b f(x) c(x) \, dx = \int_a^b f''(x) c_2(x) \, dx \ge 0 \,. $$

What I am looking for is a proof of the conjecture (the “if” direction) which works for all convex functions $f$, without additional assumptions on differentiability.

Martin R
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    Counter-example: $c: [-1,1] \to \mathbb{R}$, $c(x) = 1-3x^2$, $f(x)=x^2$ – aschepler May 01 '22 at 15:13
  • @aschepler: You are right, and my error is the integration by parts: I can assume that $c_1(a) = c_1(b) = 0$, but not that $c_2(a) = c_2(b) = 0$, so my proof is wrong if $f'(a) \ne f'(b)$ :( – Martin R May 01 '22 at 15:25

1 Answers1

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Your equivalence is not correct. A correct version appears in Problem 1.28 of Chapter 7 this book:

Let $c:[a, b] \to \Bbb R$ be a Riemann integrable function. Then $$ \int_a^b f(x) c(x) \, dx \ge 0 $$ holds for all convex functions $f:[a, b] \to \Bbb R$ if and only if $$ \int_a^b c(x) \,dx = \int_a^b x c(x) \, dx = 0 \, \text{ and } \int_a^d (d-x)c(x) dx \geq 0 \text{ for every } d\in (a,b) .$$

Here is the solution given for sufficiency (concise, as is customary in this book):

A convex, piecewise affine function is the convex combination of an affine function and functions of the form $x\mapsto \max(0,d-x)$ with $d\in (a,b)$.

Gabriel Romon
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  • Thanks, that makes a lot of sense. $c_2(x) = \int_a^x (x-t)c(t) dx$ is a second anti-derivative of $c$, and for $f \in C^2$ one has $\int_a^b f(x) c(x) dx = \int_a^b f''(x) c_2(x) dx \ge 0$. – Martin R May 03 '22 at 07:17