4
$\begingroup$

Let $f$ be a continuous function on $[a,b]$ ( $f: [a,b]\to \mathbb R$, $a < b$) such that $\int_a^b f(x) \, dx = \frac{b^2-a^2}{2}$. How can we prove that $f$ has a fixed point in $(a,b)$ without using Rolle’s theorem? (With Rolle’s theorem, we can apply it to the function $g$ defined by $g(x) = \int_a^x f(t) \, dt - \frac{x^2 - a^2}{2}$).

Addition

We can use the first mean value theorem for definite integrals applied to $ g(x)=f(x)-x $, which guarantees the existence of $ c \in ]a,b[ $, such that $ g(c)=\int_a^b g(x) \, dx=0 $.

$\endgroup$
5
  • $\begingroup$ Consider the average value of $f$ on $[a,b]$ $\endgroup$
    – MPW
    Apr 12 at 22:17
  • 1
    $\begingroup$ Note that $\int_a^b (f(x)-x) dx = 0$, so there must be a point such that $f(x)-x=0$. $\endgroup$
    – copper.hat
    Apr 12 at 22:23
  • $\begingroup$ @MPW The average value of is $\frac{a+b}2$ and what ? $\endgroup$
    – Noname
    Apr 12 at 22:26
  • $\begingroup$ I added the clarification that ( f ) takes real values. @copper.hat why? $\endgroup$
    – Noname
    Apr 12 at 22:30
  • $\begingroup$ If it is continuous and $f(x) \neq x$ for all $x, what can you say about the integral? $\endgroup$
    – copper.hat
    Apr 12 at 22:55

2 Answers 2

6
$\begingroup$

Suppose $f(x)=x$ never happens on $[a,b]$. Then, since $f$ is continuous, the graph/plot of $f(x)$ lies either strictly above $y=x$, or strictly below it.

In the first case we have $$\int_a^b f(x)~ dx \gt \int_a^b x ~dx=\frac{b^2-a^2}2.$$

A contradiction. The same in the second case.

$\endgroup$
0
2
$\begingroup$

"Without Rolle's theorem" is a bit of a sketchy and ill-defined condition to me. For instance, it's just a special case of the derivative mean value theorem.

But alternatively, somewhat suggestive of the integral mean value theorem, note that $f$ will have average value $$ \frac{b^2-a^2}{2} \frac{1}{b-a} = \frac{a+b}{2} $$ across the interval. If $f$ has no fixed point, and is continuous, then $f(x)<x$ (or $f(x)>x$) across the entire interval. But then, in the former case, $f$ at worst will be the function $$ f(x) = \begin{cases} \displaystyle x - \varepsilon(x), & \displaystyle x \in \left[a,\frac{a+b}{2} \right] \\ \displaystyle \frac{a+b}{2}, & \displaystyle x \in \left[ \frac{a+b}{2}, b \right] \end{cases} $$ for some function $\varepsilon(x) > 0$ that tends to $0$ as $x \to \frac{a+b}{2}$. (Basically it's very close to $x$, but just a bit away, but I'm also trying to maintain continuity here. You can do this other ways if you prefer.) But clearly, then, on the former interval $f(x) < \frac{a+b}{2}$, so $\int_a^b f \ne \frac{b^2-a^2}{2}$.

You can probably fill in the details from here on the other case, but that is ultimately one idea you can play with. A rough Desmos demo is here, though I'm a bit lazy and only have it working for cases where $0 \le a \le b$ or $a \le b \le 0$ (i.e. not $a \le 0 \le b$). The red regions ultimately should correspond to the missing area, in the case $0 \le a \le b$ and $f(x)<x$.

enter image description here


copper.hat's comment is a lot simpler and cleverer and I'm saddened I didn't notice it. Simply note that $$ \int_a^b x \, dx = \frac 1 2 x^2 \bigg|_a^b = \frac{b^2-a^2}{2} = \int_a^b f(x) \, dx $$ so $$ \int_a^b \Big( f(x) - x \Big) \, dx = 0 $$ But then, since $f$ and $x$ are continuous, this means that $f(x)-x=0$ at some point in $[a,b]$. (If $f(x_0)-x_0 > 0$, then it is positive in a neighborhood of $x_0$, contributing positively to the integral -- but then it needs to be negative in a neighborhood of some other point to cancel that out, and then continuity gives the claim. Or $f(x)=x$ everywhere, making the claim trivial. You can fill in the details.)

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .