Task is:

$f(x)$ is positive, continious function in the field of real numbers, and $\int_{-\infty}^{\infty}f(x)dx=1$. Let $\alpha\in(0,1)$, and length of $[a,b]$ is minimal from all $\int_{a}^{b}f(x)dx=\alpha$. Prove that $f(a)=f(b)$

I've tried to use mean value theorem.

And I've tried to use some knowledge from probability theory, because $f(x)$ is seems like probability density function.

But for now, I haven't came any closer to proof.


The trick is reframing it in language that is familiar to optimization.

Let $g(a,b) = b - a$. We want to minimize $g$ subject to the constraint $h(a,b) = \int_a^b f(t) \ dt = \alpha$ for some $\alpha \in (0,1)$.

Note that the constraint isn't vacuous because $\int_{\mathbb R}f = 1 $ implies there must be at least one pair $(a,b)$ which satisfies $h(a,b) = \alpha$.

Using now Lagrange multipliers, $\nabla g - \lambda \nabla h = 0$ iff

$$-1 + \lambda \frac{\partial h}{\partial a} = 0 \ \ \ \text{ and } \ \ \ 1 + \lambda \frac{\partial h}{\partial b} = 0$$

Apply the Fundamental Theorem of Calculus and you're done.

This is a nice result. I tried at first to construct a counterexample.

Intuitively, I think I've convinced myself it makes sense: on the optimizing interval $[a,b]$, there is some value of $x \in (a,b)$ for which $f(x) > \max\left( f(a), f(b) \right)$. Now look at alternative scenario intervals $J= [a\pm\delta_1, b\pm\delta_2]$, which maintain $\int_J f = \alpha$. If $f(a) \neq f(b)$ it looks like we can make $J$ shorter than $b - a$.

If anyone can turn that into a formal argument it would be interesting to see.

  • $\begingroup$ Very roughly: if f(a)>f(b) then shifting the region [a,b] left by δ would increase the integral by approximately δ(f(a)-f(b)) which would then allow you to shrink the length of the interval and still achieve the same integral. $\endgroup$ – Dan Piponi Mar 12 '15 at 0:19
  • $\begingroup$ Thank you, I'll think this through and mark as resolved! $\endgroup$ – DoctorMoisha Mar 12 '15 at 10:03

Making precise the intuition that has been suggested by Simon S.

Assume that $f(b)>f(a)$. Then from continuity of $f$ we can find a $\delta$ such that $f(x_1)< f(x_2) $ $\forall x_1 \in [a,a+\delta],\,\forall x_2 \in [b,b+\delta]$. The previous implies that $\int _a^{a+\delta} f(x)d x<\int _b^{b+\delta} f(x)d x\fbox{1}$.

Now define the continuous function $F(x)=\int _{a+\delta}^x f(x)d x$.

Then since $f$ is always positive $F(b)=\int _{a+\delta}^b f(x)d x<\alpha$. Then again from $\fbox{1}$ we see that $F(b+\delta)=\int _{a+\delta}^{b+\delta} f(x)d x=\int _{a}^{b} f(x)d x+ \int _b^{b+\delta}f(x)d x- \int _a^{a+\delta} f(x)d x>\alpha$. Hence from continuity of $F$ there is a $c<\delta$ such that $F(b+c)=a$.

But $b+c-a-\delta<b-a$ a contradiction.


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.