Mean value theorem with integrals?

Question

Here is the question:

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

I'm pretty sure that the problem wants me to use the mean value theorem of some type. So I tried to consider a function $F(t)$ that would give the form $$F'(t) = \int_0^t xf(x)\,dx -tf(t)$$ so that I would be able to say $F'(c) = 0$ for some $c\in (0, 1)$, using Rolle's theorem. But this gave me $$F(t) = \int_0^t ((t-1)x-x^2)f(x)\,dx$$ which didn't really help me proceed any further.

I have also tried setting $$F(t) = \int_0^t (t-x)f(x)\,dx$$ in hopes of using Rolle's theorem, since $F(0)=F(1)=0$. But $F'(t)$ wasn't really the required form. I've also tried other different forms so that I could apply mean value theorem for integrals, or Cauchy's mean value theorem. But I couldn't derive the correct form to solve the problem.

Maybe I have missed something? Or can someone provide me a different approach to this problem? Thanks in advance.

Answer 1

Define $F, G : [0,1] \to \mathbb{R}$ as $F(x) = \displaystyle\int_{0}^{x}f(t)dt$ and $G(x) = \displaystyle \int_{0}^{x}tf(t)dt.$ Integrating the second integral by parts, allows us to write:

$G(x) = xF(x) - \displaystyle \int_{0}^{x}F(t)dt.$

Note that $G(0) =0.$ Our aim is to find another zero of $G$, say $b$, and then apply Rolle's theorem to the function $e^{-x}G(x)$ on the interval $[0, b].$

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

Proof of claim: Suppose not. Then since $G$ is continuous, it cannot change sign on $(0,1]$ so WLOG, assume $G(x) >0 \quad \forall \, x >0.$

Since $F$ is continuous on a closed and bounded interval $[0, 1]$, it attains its bounds. Let $d \in [0,1]$ be a point of minimisation of $F.$ Two cases arise:

1) $d=0$

Then $\forall \, t \in [0,1] \quad F(t) \geq F(0) =0.$ By the mean value theorem for integrals, $\, \exists \, 0< x_{0} < 1$ such that $\displaystyle \int_{0}^{1}F(t)dt = F(x_{0}).$ The given condition can be stated as $\displaystyle \int_{0}^{1}F(t)dt =0$, hence $F(x_{0}) =0.$

By assumption, $G(x_{0})>0$ which implies $\displaystyle \int_{0}^{x_{0}}F(t)dt< 0$ which by the mean value theorem again implies that $x_{0}F(x_{1})< 0$ for some $x_{1} \in (0, x_{0})$ and thus $F(x_{1}) <0, $ a contradiction. Thus $d=0$ is an impossibility and hence $d>0$ which brings us to case 2.

2) $d>0$

For all $t \in [0,1]$ we have $F(t) \geq F(d).$ Integrating over $[0,x]$ we get $\displaystyle \int_{0}^{x}F(t)dt \geq xF(d)$ and rearranging this yields $G(x) \leq x(F(x) - F(d))$ from which we get that $G(d) \leq 0$ for $d>0,$ a contradiction. Hence the claim holds $\blacksquare$.

With the claim in hand, we apply Rolle's theorem to the function $h(x) = e^{-x}G(x)$ on the interval $[0, b]$ to get $c \in (0, b)$ such that $h'(c) =0$ which implies $e^{-c}(G'(c) - G(c))=0$ from which it follows that $\displaystyle \int_{0}^{c}tf(t)dt = cf(c).$

Answer 2

The very first integral can be splited in two. Value of one integral is f(c), according to L. M. V. T. and of other is c. f(c). Comparing c. f(c) =f(c), we get c=1 or f(c) =0.But c cannot be 1. So f(c) =0.

Answer 3

In fact, the conclusion is true for any continuous function $f$ that changes its sign in $[0,1]$.

Let $G(y)=\int_0^{y}xf(x)dx$, and assume that there is no point $y$ such that $G'(y)=G(y).$

I claim that there can't exist a point $b>0$ where $G'(b)<G(b)$ and $G(b)>0$. Indeed, let $b$ be such a point, and let $a:=\inf\{0<x<b:G>0 \text{ on } (x;b) \}$. Then, on $(a,b)$, we have $(\log G(x))'= G'(x)/G(x)<1$, hence $\log G(x)> (x-b)+\log G(b)$ and thus $G(x)> G(b)e^{x-b}\geq G(b)e^{-b}$. On the other hand, $G(a)=0$, regardless of whether $a=0$ or $a>0$, which is a contradiction. Similarly, there can't exist a point $b>0$ with $G(b)<0$ and $G'(b)>G(b)$.

Hence, if $a$ is some point where $G(a)>0$, then we necessarily have $G'(y)>G(y)>0$ in some interval $(a,b)$. If take $b$ to be the supremum of all possibilities, then we still have $G(b)>G(a)>0$, and hence we can only have $b=1$. Similarly, if $a$ is such that $G(a)<0$, then in fact $G'(y)<G(y)<0$ for all $y\in [a;1]$. But these scenarios are mutually exclusive. On the other hand, $\inf\{x:G(x)\neq 0\}=0$, for otherwise $G'\equiv G\equiv 0$ in some neighborhood of $0$.

Therefore, either $xf(x)=G'(x)>G(x)>0$ for all $x\in (0,1]$, or $xf(x)=G'(x)<G(x)<0$ for all $x\in (0,1]$. In each case, $f$ has a constant sign.

Answer 4

You can use Taylor Lagrange equality between expansion :

I set $$ F(t)\triangleq \int_{0}^t(1-x)f(x)dx$$

So as $F(t)$ is differentiable , by Lagrange there exists $c \in ]0;1[$

So that $$ F(1)=F(0)+F'(c)$$

Which respectively gives :

$$ cf(c)=f(c) $$ so $f(c)=0$ or $c=1$. Yet $c$ isn't equal to $1$ according to the theorem,

So know, you have to proove

$$ \int_0^cxf(x)dx=0$$