Let $f$ be continuous on $[a,b]$, and differentiable on $(a,b)$, $\int_a^b f(x)dx=0$. Show that there exists two distinct $\xi,\eta\in(a,b)$, such that $$f'(\xi)(\xi-a)+f'(\eta)(\eta-b)+f(a)+f(b)=0$$.

My attempt: Let $F(x)=(x-a)[f(x)+f(a)]-\int_a^x f(t)dt$, $G(x)=(x-b)[f(x)+f(b)]-\int_b^x f(t)dt$. Then $F(a)=0, F(b)=-G(a), G(b)=0.$ And what we need is to show some $\xi\neq \eta$, such that $F'(\xi)+G'(\eta)=0$. Consider $0<c<1$, and by the Lagrange intermediate value, it suffices to show for some $c$, $\frac{F(c)}{c-a}+\frac{-G(c)}{b-c}=0$. But such a $c$ is not obvious by continuity.

  • $\begingroup$ Maybe by using a two-steps procedure like in this interesting problem and solution: math.stackexchange.com/q/3715320 but I am not able to find the good auxiliary function... $\endgroup$
    – Jean Marie
    Jun 11 at 10:05

The statement is wrong, a counterexample is $f(x) = x^2-1/3$ on $[a, b]= [-1, 1]$.

The condition $ \int_{-1}^1 f(x) \, dx = 0$ is satisfied, but for all $\xi, \eta \in (-1, 1)$ is $$ f'(\xi)(\xi-a)+f'(\eta)(\eta-b)+f(a)+f(b) \\ = 2\xi(\xi+1) + 2\eta(\eta-1) + \frac 43 \\ = 2 \left(\xi+\frac 12\right)^2 + 2\left(\eta-\frac 12\right)^2 + \frac 13 \ge \frac 13 > 0 \, . $$

Some remarks on how I came up with this example. Of course I tried to prove the statement, and wanted to determine $\xi$ and $\eta$ for some concrete cases. In order to make things simple, I chose $[a, b] = [-1, 1]$.

If $f$ is an odd function then $\eta = -\xi$ always works: $$ f'(\xi)(\xi+1)+f'(\eta)(\eta-1)+f(-1)+f(1) \\ = f'(\xi)(\xi+1)+f'(\xi)(-\xi-1) = 0 \, . $$

The “simplest” even functions on $[-1, 1]$ are $f(x) = x^2+c$, and with $c=1/3$ the condition $ \int_{-1}^1 f(x) \, dx = 0$ is satisfied. Then I tried to find a solution of the form $\eta = -\xi$, but that gave a quadratic equation without real solutions.

This lead to the conjecture that the statement might be wrong, and that $f(x) = x^2-1/3$ could be a counterexample.

  • $\begingroup$ I was thinking about this problem since it was asked. Great result! Thanks for sharing. Will give the bounty as soon as the site allows. $\endgroup$
    – VIVID
    Jun 12 at 18:49
  • $\begingroup$ @VIVID: So was I :) – Thanks! $\endgroup$
    – Martin R
    Jun 12 at 18:51
  • 1
    $\begingroup$ Shout out for sharing your working process to find the example. Very insightfull. Great answer. $\endgroup$
    – Cornman
    Jun 12 at 19:36
  • $\begingroup$ (+1) Well done! $\endgroup$
    – Mark Viola
    Jul 13 at 16:08

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