Simple Solution of a Functional Equation?

Let us define $h(x)=\frac{x}{x-1}.$ It is easy to see that $h^{-1}=h,$ and $h((1, 2])=[2, \infty)$ , $\;h([0, 1))=(-\infty,0],$ $\;h((\infty,0])=[0, 1),$ and $h([2, \infty))=(1, 2].$ We are going to start constructing a solution to the given functional equation. First, we are going to define a function $g$ by the following: $g(x) = \left\{ \begin{array}{ll} x, & \quad 0\leq x \leq 1 \\ 2-x, & \quad 1\leq x \leq 2 \\ 0, & \quad x<0\;\text{or}\; x>2 \end{array} \right.$

Now, we can define a solution for the given functional equation in the following form: $f_0(x) = \left\{ \begin{array}{ll} \int_{-\infty}^x g(t)dt, & \quad 0\leq x \leq 2 \\ f_0(h(x)), & \quad x<0\;\text{or}\; x>2. \end{array} \right.$

It is easy to verify that the function $f_0$ defined in that way satisfies the functional equation. To prove that it is differentiable at any real number, we need to prove only that it is differentiable at 0 and 2. We are going to prove, for example, that it is differentiable at 2 and then the proof of the differentiability at 0 will be similar. Using the Fundamental Theorem of Calculus, we can see that $f'(2-0)=0.$ Now let us prove that $f'(2+0)=0.$ $f_0'(2+0)= \lim_{x\rightarrow 2+}\frac{f_0(x)-f_0(2)}{x-2}= \lim_{x\rightarrow 2+}\frac{f_0(h(x))-f_0(2)}{x-2}= \lim_{x\rightarrow 2+}\frac{f_0(h(x))-f_0(2)}{h(x)-2}\dot\frac{h(x)-2}{x-2}= \lim_{z\rightarrow 2-}\frac{f_0(z)-f_0(2)}{z-2}\lim_{x\rightarrow 2+}\frac{h(x)-2}{x-2}=0*h'(2)=0.$ Then $f_0'(2)$ exists and is equal to 0.

In a similar way, we can prove that $f_0'(0)$ exists and is equal to 0. Therefore, we have proved that the function $f_0$ is a differentiable solution of the given functional equation and its derivative has exactly two distinct real zeros. Notice that $f_0$ is monotone on each one of the intervals $(-\infty, 0), (0, 2)$ and $(2, \infty)$ . So the functional equation has differentiable solutions for which the minimum number of real zeros has to be less than or equal to 2.

Now, we are going to prove that for every differentiable solution of the functional equation, its derivative must have at least two distinct real zeros. Notice that for whatever solution of the functional equation $f(1.5)=f(3)$ and $f(0.5)= f(-1),$ and, therefore, using Rolle's Theorem , its derivative $f'$ has at least two real zeros in the intervals $(-1, 0.5)$ and $(1.5, 3),$ respectively, that, of course, will be distinct. So the answer to the question is that the functional equation has differentiable solutions and the minimum number of real zeros for the derivative of a solution is 2.

Without loss of generality, we can assume that $f(0) = 0$ . If we define the function $g(x) = f(x^{-1})$ for $x \neq 0$ then it follows that $g$ is differentiable for $x \neq 0$ and that $g(x) = g(1-x)$ for $x \neq 0,1$ . Thus $\lim_{x \to 0}g(x) = \lim_{x \to 1}g(x) = g(1)=f(1)$ and so, if we define $g(0) = f(1)$ , then $\lim_{x \to0}\frac{g(x) - g(0)}{x} \; =\; \lim_{x \to 0}\frac{f(1-x) - f(1)}{x} \; = \; -f'(1)$ so that $g$ extends to a differentiable function $G$ on $\mathbb{R}$ such that $G(x) = G(1-x)$ for all $x \in \mathbb{R}$ and $f(x) = G(x^{-1})$ for $x \neq 0$ . But then $\begin{aligned} \lim_{x \to \infty} G(x) & = \; \lim_{x \to \infty}f(x^{-1}) \; = \; \lim_{x \to 0}f(x) \; = \; f(0) \; = \; 0 \\ \lim_{x \to \infty}xG(x) & = \; \lim_{x \to \infty}\frac{f(x^{-1})}{x^{-1}} \; = \; \lim_{x \to 0}\frac{f(x)-f(0)}{x} \; =\; f'(0) \end{aligned}$ But since $xG(x) \; = \; xG(1-x) \; = \; G(1-x) - (1-x)G(1-x)$ we deduce that $\lim_{x \to \infty}xG(x) = -\lim_{x \to\infty}xG(x) = 0$ , and hence $f'(0) = 0$ . Since $G(x) = f(x^{-1})$ , we see that $G'(x) = -x^{-2}f'(x^{-1})$ for $x \neq 0$ , and so $-4f'(2) = G'(\tfrac12) = -G'(\tfrac12) = 0$ . Thus $f'$ vanishes at $0$ and $2$ .

We deduce that $f$ exists, and that $f'$ has at least $2$ zeros. A particular example of $f$ is $f(x) \; = \; \left\{ \begin{array}{lll} \mathrm{sech}\,(x^{-1} - \tfrac12) & \hspace{1cm} & x \neq 0 \\ 0 & & x = 0 \end{array} \right.$

Given , f(x)=f(x/x-1) Now let t=1/(x-1) (for x≠1 and t≠0)
Therefore f(1+t)=f(1+1/t) Differentiating both sides , f'(1+1/t)(-1/t^2)=f'(1+t) =>f'(1+t)+f'(1+1/t)/t^2=0 For t=1/t f'(1+t)=f'(1+1/t) Thus t=±1 Therefore f'(0)(1+1/1) = 0 and f'(2)(1+1/1)=0 Therefore f'(0)=f'(2)=0 So f'(x) will have minimum two zeroes.