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I have a Lebesgue integrable function $f : [0,1] \to [0,1]$ that solves the equation

$$ 4 x f(x^2 ) = f(x) + f(1-x)$$ for all $x \in [0,1]$. Is it possible to give an analytic expression for $f$ ?

Edit 1: Is there a non-trivial solution $f \neq 0 $?

Edit 2: I forgot to noticed that the function $f$ is Lebesgue integrable. I added it now.

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3 Answers 3

up vote 3 down vote accepted

Suppose that $f$ is continuous, and not the zero function, and put $M={\rm Sup}\{f(x), x\in [0,1]\}$. There exists $x_0$ such that $M=f(x_0)>0$. Taking $x=\sqrt{x_0}$, we get $\displaystyle 4\sqrt{x_0}f(x_0)=f(x)+f(1-x)\leq 2M$, hence $\displaystyle 4\sqrt{x_0}\leq 2$, and $\displaystyle x_0\leq \frac{1}{4}$. By the functional equation with $x=x_0$, we get: $$M+f(1-x_0)=f(x_0)+f(1-x_0)=4x_0f(x_0^2)\leq f(x_0^2)\leq M$$ Hence $f(1-x_0)\leq 0$, and $f(1-x_0)=0$. This imply $4x_0=1$ and $f(x_0^2)=M=f(1/16)=f(1/4)$. Now taking $\displaystyle x_1=1/16$ in the functional equation:

$$f(1-x_1)=4x_1f(x_1^2)-f(x_1)=4x_1f(x_1^2)-M\leq 4x_1M-M=-3M/4<0$$ a contradiction.

EDIT:

A new solution, where I do not suppose $f$ continuous. Put $M={\rm Sup}\{f(x)\}$ ($M$ exists as $f$ is from $[0,1]$ to $[0,1]$), and suppose that $M>0$.

There exists $x_n$ such that $f(x_n)\to M$. Now let $u$ any value such that there exists a subsequence $x_{n_k}$ of $x_n$ such that $x_{n_k}\to u$ (we have used the compacity of $[0,1]$). Let $y_k=\sqrt{x_{n_k}}$. We get $\displaystyle 4\sqrt{x_{n_k}}f(x_{n_k})=f(y_k)+f(1-y_k)\leq 2M$, and hence $4\sqrt{u}M\leq 2M$. We get $u\leq 1/4$.

Now $\displaystyle f(1-x_{n_k})= 4x_{n_k}f(x_{n_k}^2)-f(x_{n_k})$. Take a subsequence of $x_{n_k}$ say $x_{n_{k_p}}$ such that $f(x_{n_{k_p}}^2)$ is convergent, say to $L\leq M$. We get that $4x_{n_{k_p}}f(x_{n_{k_p}}^2)-f(x_{n_{k_p}})$ is convergent to $4uL-M$. Now if $L<M$ or $4u<1$, $4uL-M<0$ and for $p$ large, we get $f(1-x_{n_{k_p}})<0$, a contradiction. We have thus $L=M$ and $u=1/4$. Now put $y_p=x_{n_{k_p}}^2$. We have $f(y_p)\to M$. By what we have show, there exists a subsequence of $y_p$ that converge to $1/4$. But $y_p$ is convergent to $1/16$, contradiction.

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Thank you! Really nice. –  Adam Aug 24 at 16:10
    
After thinking about it - why should the function attain its supremum? I mean so far there are no conditions on f right? –  Adam Aug 26 at 16:28
    
@Adam I think that I have taken the function continuous, but apparently you have not this hypothesis... I edit my answer.Thank you for the remark. –  Kelenner Aug 26 at 16:29
    
@Adam: See the new solution, and say to me if it is correct. –  Kelenner Aug 26 at 17:54
1  
@Adam My English is bad, so I did not understand your comment, in particular for the last line of my answer. In fact I have said that a subsequence of $y_p$ is convergent to $1/4$, not that $y_p$ converge to $1/4$. (In fact, we start with a sequence $x_n$ such that $f(x_n)\to M$, and we show that every convergent subsequence of $x_n$ is convergent to $1/4$; then $x_n$ is really convergent to $1/4$. As $f(y_p)\to M$, we have $y_p\to 1/4$) You can use absolutely freely the material here. It was a very interesting question. Good continuation, and good luck for your paper. –  Kelenner Aug 26 at 20:03

Here's one solution: $f(x) = 0$ for all $x$.

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Thanks for the solution. I edited my question now, because I wanted to find a non-trivial solution. –  Adam Aug 24 at 14:15
    
If I'm right the non-trivial solution is the only solution –  Mr.T Aug 24 at 14:48

$4xf(x^2)=f(x)+f(1-x)$ means $2f(x^2)d(x^2)=f(x)dx-f(1-x)d(1-x)$. Put $F(x)=\int_0^x f(t)dt\ge 0$ then:

$$F(x)=2F(x^2)+F(1-x)\ge 2F(x^2)$$

which means $\forall n\in\mathbb{N}$: $F(x)\ge 2^nF(x^{2^n})=2^nF(1-x^{2^n})+2^{n+1}F(x^{2^{n+1}})\ge2^nF(1-x^{2^n})$ or $F(1-x^{2^n})\le\frac{F(x)}{2^n}$. Since $F$ is continous and nonnegative, taking the limit when $n\rightarrow\infty$ we must conclude that $F(1)=0$ which means $f(x)=0$ $\forall x\in[0,1]$

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I have found $2F(x^2)=F(x)-F(1-x)+\int_0^1 f(t)dt$. Can you verify ?. (This is your equality when $\int_0^1 f(t)dt=0$). Thanks in advance. –  Kelenner Aug 24 at 15:22
    
I just integrate the first line in my answer... How did you get that. I think it's wrong... –  Mr.T Aug 24 at 15:30
2  
Putting $u=1-t$, I have: $$\int_0^x f(1-t)dt=-\int_1^{1-x}f(u)du=-F(1-x)+\int_0^1f(u)du$$ –  Kelenner Aug 24 at 15:42
    
Ok it seems like I have made a fatal error. Thanks –  Mr.T Aug 24 at 15:51

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