# Show that there is a $c\in(0,1)$ such that $f(c)=\int_0^cf(x)dx$.

Question: Let $$f:[0,1]\to\mathbb{R}$$ be a continuous function such that $$\int_0^1f(x)dx=\int_0^1xf(x)dx.$$ Show that there is a $$c\in(0,1)$$ such that $$f(c)=\int_0^cf(x)dx.$$

My solution: Define the function $$g:[0,1]\to\mathbb{R}$$, such that $$g(x)=x\int_0^x f(t)dt-\int_0^x tf(t)dt, \forall x\in[0,1].$$

Now since $$f$$ is continuous on $$[0,1]$$, thus we can conclude by Fundamental Theorem of Calculus that $$g$$ is differentiable $$\forall x\in[0,1]$$ and $$g'(x)=\int_0^x f(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$

Observe that $$g(0)=g(1)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists b\in(0,1)$$, such that $$g'(b)=0$$, i.e $$\int_0^b f(t)dt=0.$$

Now define $$h:[0,1]\to\mathbb{R}$$, such that $$h(x)=e^{-x}g'(x), \forall x\in[0,1].$$

Now $$h'(x)=-e^{-x}g'(x)+g''(x)e^{-x}=e^{-x}(g''(x)-g'(x)), \forall x\in[0,1].$$

Observe that $$h(0)=h(b)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists c\in(0,b)\subseteq (0,1)$$, such that $$h'(c)=0$$. This implies that $$e^{-c}(g''(c)-g'(c))=0\\\implies g''(c)-g'(c)=0\hspace{0.3 cm}(\because e^{-c}\neq 0)\\\implies f(c)=\int_0^cf(x)dx.$$

Is this solution correct? And is there a better solution that this?

• Your solution looks fine. Apr 16 '20 at 5:18
• Isn't this something like the reverse Lagrange's mean value theorm? Correct me if I'm wrong... Apr 16 '20 at 6:36
• See related question math.stackexchange.com/q/3610557/72031 Apr 16 '20 at 7:55

Your proof is correct. This is another one.

We may assume that $$f$$ is not identically zero (otherwise it is trivial). Since $$f$$ is continuous and $$\int_0^1(1-x)f(x)\,dx=0$$ we have that $$M=\max_{x\in [0,1]}f(x)>0$$ and $$m=\min_{x\in [0,1]}f(x)<0$$. Moreover $$\exists x_M,x_m\in [0,1]$$ such that $$f(x_M)=M$$ and $$f(x_m)=m$$. Let us consider the following continuous map $$F(x):= f(x) - \int_0^xf(t)\,dt.$$ If $$x_M<1,$$ then $$F(x_M)=M-\int_0^{x_M}f(t)\,dt\geq M- Mx_M >0.$$ If $$x_M=1$$ then, $$F(x_M)=M-\int_0^{1}f(t)\,dt> 0$$ because $$f$$ strictly less than $$M$$ in an interval of positive length containing $$x_m$$. In both cases we conclude that $$F(x_M)>0$$. In a similar way, we show that $$F(x_m)<0$$.

Finally, since $$F$$ is continuous on $$[0,1]$$, it follows, by the Intermediate Value Theorem, that there exists $$c$$ strictly between $$x_M$$ and $$x_m$$, and therefore $$c\in (0,1)$$, such that $$F(c)=0$$, that is $$f(c)=\int_0^cf(t)\,dt.$$

• +1 It appears that the approach would also work for the question linked in my comment to the current question. Apr 23 '20 at 2:02
• Another observation: what you have proved is that if a continuous $f$ changes sign in $[0,1]$ then we have a $c\in(0,1)$ such that $f(c) =\int_{0}^{c}f(x)\,dx$. Apr 23 '20 at 2:08

As noted by RobertZ, your proof is correct. Here is another proof that follows the same outline as yours: first we find another zero for the antiderivative of $$f$$ and then we use Rolle's theorem in an appropriate way. This approach is admittedly more long-winded but doesn't make use of the $$e^{-x}$$ trick.

Define $$F: [0,1] \to \mathbb{R}$$ as $$F(x) =\displaystyle \int_{0}^{x}f(t)dt.$$ Note that the given condition can be stated as $$\displaystyle \int_{0}^{1}F(t)dt =0$$

Claim 1: There exists $$b \in (0,1)$$ such that $$F(b) =0.$$

Proof of claim 1: By the mean value theorem for integrals there exists $$b \in (0,1)$$ such that $$F(b)= \displaystyle \int_{0}^{1}F(t)dt$$, which implies $$F(b)=0.$$

Now, we look for an appropriate sub-interval of $$[0,b]$$ on which we can apply Rolle's theorem to $$g.$$

Let $$G(x)=\displaystyle \int_{0}^{x}F(t)dt$$ and define $$g:[0,b] \to \mathbb{R}$$ by $$g(x)= G(x) -F(x).$$

Claim 2: $$g$$ is not injective on $$[0, b].$$

Proof of claim 2: Suppose not. Then $$g$$ is injective and since it is clearly continuous too, $$g$$ is monotone. WLOG, let $$g$$ be monotone increasing. Then since $$g$$ is differentiable, $$g'(x) \geq 0 \, \forall \, x \in [0,1].$$ If there exists at least one $$x$$ for which $$g'(x) =0$$ we are done so assume $$g'(x)>0.$$ Since $$g(0) =0,$$ we have $$g(x)>0$$ for all $$x \in (0,b].$$

Let $$x_{0}$$ be a point of maximisation for $$F.$$ Assume $$F$$ is not identically $$0$$ or else $$f$$ is and the problem is trivial. We claim that there exists $$c \in (0, b)$$ such that $$F(c)<0.$$ If $$x_{0}=0$$ or $$b$$ then $$F\leq 0$$ so if $$F$$ is not identically $$0$$ choose an other point of $$(0, b)$$ to be $$c.$$ If $$x_{0} \in (0, b)$$ then since $$g({x}_{0})>0, F(x_{0})< \displaystyle \int_{0}^{x_{0}}F(t)dt \leq x_{0}F(x_{0}).$$

If $$F(x_{0}) \neq 0$$ we get $$x_{0} \geq 1,$$ a contradiction. Hence $$F(x_{0})=0$$ and since $$F$$ is not identically $$0$$ there exists some $$c \in (0, b)$$ such that $$F(c)<0.$$

Since $$F$$ is a continuous function on a closed and bounded interval $$[0, b]$$, it attains its bounds. In particular $$\exists \, d \in [0, b]$$ such that $$F(d)\leq F(x) \, \forall \, x \in [0,b].$$ Clearly $$d\neq 0, 1$$ or else $$F(x) \geq 0 \, \forall x \in [0,b]$$ contradicting the fact that $$F(c) <0.$$ Therefore $$d \in (0,b)$$ and since it is a point of minimisation, $$F'(d) =0.$$ Then $$F(d)= F(d) -F'(d) =g'(d)>0> F(c)$$ contradicting the fact that $$d$$ is a point of minimisation of $$F.$$ Therefore our hypothesis that $$g$$ is injective is false and hence $$g$$ is not injective and there exists $$a, a' \in [0, b]$$ with $$a \neq a'$$ such that $$g(a) =g(a').$$

Then since $$g$$ restricted to $$[a, a']$$ satisfies the conditions for Rolle's Theorem, there exists some $$x_0 \in (a,a')$$ such that $$g'(x_0)=0$$ which implies $$F(x_0)=F'(x_0)$$ from which it follows that $$f(x_{0}) = \displaystyle \int_{0}^{x_{0}}f(x)dx$$

Note that the proof follows almost identically if we assume $$g$$ to be monotone decreasing in the proof of the claim.