# Fixed point for some continuous function.

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 $$g(x) = \int_a^x f(t) \, dt - \frac{x^2 - a^2}{2}$$).

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 $$g(c)=\int_a^b g(x) \, dx=0$$.

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

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.

"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 $$f(x) (or $$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).

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.)