9
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
11
+50
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.