# A functional equation while attempting to integrate a function differently: $f(x)-f(y) = z \bigl( f(xz) - f(yz) \bigr)$

Here is a functional equation :

For $$f : \mathbb (0,\infty) \to \mathbb R$$ continuous non-constant, $$f$$ has the property that for every $$b,x,y>0$$, $$f(1) = -1$$ and $$f(x) - f(y) = b \bigl( f(bx) - f(by) \bigr) \text .$$

I want to show that the only possible solutions are $$f(x) = a-\frac {a+1}x$$ for some constant $$a$$, without using any form of calculus(with the exception of limits, if required).

For an approach, there are too many free variables, so I tried to set $$b$$ to a certain value, say $$b=2$$. Then I get: $$f(x)-f(y) = 2 \bigl( f(2x) - f(2y) \bigr) \text .$$

but I don't know how to proceed : I can't use this to find a particular value of any single $$x$$, because the value of $$x$$ depends on the value of $$2x$$, and so on, so there's no way of proceeding I can think of. This also makes me feel that there is missing information, but that doesn't seem likely because in the presence of three constraining variables and domain, some kind of uniqueness must hold, right?

Setting $$g(x) = xf(x)$$ and trying to prove that it is constant has also proved futile, for the same reason as before.

Context : This post contains the question I am trying to answer.

Let $$f : (0,\infty) \to \mathbb R$$ be a function satisfying $$f(x)-f(y) = \int_y^x \frac{\mathrm dt}{t^2}$$ for all $$y,x>0$$. Under the fix $$f(1) = -1$$, we know that there is a unique function $$f$$ satisfying these conditions, given by $$f(x) = -1 + \int_1^x \frac{1}{t^2}\,\mathrm dt = -\frac 1x$$. I attempted to locate such an $$f$$ without using calculus, but rather using functional equations.

The logic was as follows : let $$b>0$$ be arbitrary, and in the expression $$\int_x^y \frac{\mathrm dt}{t^2}$$, make the change of variables $$bt =u$$, so that $$b\,\mathrm dt = \mathrm du$$. Accordingly, we get: $$f(x)-f(y) = \int_x^y \frac{\mathrm dt}{t^2} = \int_{bx}^{by} \frac{b^2 \,\mathrm du}{u^2} = b \int_{bx}^{by} \frac{\mathrm du}{u^2} = b \bigl( f(bx)-f(by) \bigr) \text .$$

and thus the functional equation comes out. Now if I can use non-calculus techniques to prove that this functional equation is satisfied only by $$-\frac 1x$$ then I will have obtained the antiderivative of $$\frac 1{x^2}$$.

• I'm afraid that won't work: $\displaystyle f(x)=a-\frac{a+1}x$ (for any constant $a$) satisfies your functional equation, and $f(1)=-1$.
– user436658
Jan 22 at 20:04
• @ProfessorVector Thank you very much! There's something missing now : according to the second part of the question, I derived a functional relation that I thought was enough to decide the nature of $f$. The function that you mention is the specific case of my function for $a=0$, which gives $f(x) = -\frac 1x$. Can you prove that this is the list of all possible functions i.e. that if a function $f$ satisfies the constraints then there exists a constant $a$ such that $f(x) = a-\frac{a+1}{x}$? I will edit the question to reflect this. Jan 22 at 20:11
• Yes, my goal was definitely to get a functional equation and I tried to reduce part of the problem to algebraic techniques. You got me completely right. I love solving functional equations too much. Thanks to you, I have refreshed my math backround. I learned new things and realized my mistakes. Thanks a lot. Jan 23 at 17:12
• @lonestudent Yes, good to see you visited here. I struggled for some time to solve this functional equation, Professor's answer is brilliant. Jan 23 at 17:13
• By the way, I want you to know that I upvoted your question (and given answers) before. It doesn't mean anything, but I wanted to say. Functional equations are my favorite math problems. Jan 23 at 17:21

In this case, more free variables is better: let's assume $$f(x)-f(y) = z(f(xz)-f(yz))$$ for $$x,y,z>0$$. Then, $$f(x)-z\,f(xz)=f(y)-z\,f(yz),$$ meaning that expression depends only on $$z$$: $$f(x)-z\,f(xz)=g(z).$$ Letting $$x=1$$, we see that $$g(z)=-1-z\,f(z)$$, i.e. we have $$f(x)-z\,f(xz)=-1-z\,f(z).$$ Exchanging $$x$$ and $$z$$, we get $$f(z)-x\,f(xz)=-1-x\,f(x),$$ multiplying the first equation by $$x$$ and the second one by $$z$$ and subtracting, we arrive at $$x\,f(x)-z\,f(z)=-x+z+xz(f(x)-f(z)).$$ This equation has to hold for any $$x,z>0$$, substituting any constant value $$z_0\neq1$$ for $$z$$, we see that $$x\,f(x)=ax+b$$. Since $$f(1)=-1$$, this means $$b=-a-1$$, i.e. $$x\,f(x)=ax-a-1$$.

• Excellent, the exchange followed by the subtraction was what I was missing. Thank you so much, may you be granted myriad wishes, Professor. Jan 22 at 20:30
• For functional equations, that's a rather standard technique, but then, that's a somewhat exotic subject. Essentially, you need it only in mathematical contests. :-)
– user436658
Jan 22 at 20:35
• True, I believe functional equations as a rigorous mathematical subject go into iteration theory (which I though of because $\frac 1x$ is an involution of order $2$ so a switch of $x \to \frac 1x$ could produce something, did not happen though), Schroder's equation, then of course functional equations involve the derivative and integral so there's PDE and integro-differential equations. But yes, for contests I read up a few books, none of which had this because they were into Cauchy's equation and standard forms which I knew did not apply. Jan 22 at 21:02
• I used PARI/GP and Mathematica to check if $\,f(x):=ax-a-1\,$ is a solution of the equation $\,f(x)-f(y) = z(f(xz)-f(yz))\,$ and it failed to check. the left side is $\,a(x-y)\,$ while right side is $\,z^2a(x-y).$ Jan 22 at 22:49
• @Somos Hi, you seem to have used $f(x) = ax-a-1$ for the PARI/GP and Mathematica check, which I see from the results you obtain. Professor had $f(x) = \frac{ax-a-1}{x} = 1-\frac{a-1}{x}$, which works for any $a$ , because here $f(x) = a - \frac{a-1}{x}$, so $f(x) - f(y) = (a-1)(\frac{1}{x} - \frac 1y)$,then replacing $x,y$ by $bx,by$ just puts a $\frac 1b$ on the RHS, so this satisfies the equation. Jan 23 at 6:37

Suppose we have a functional equation $$x(f(x)-f(x\,y)) = z(f(z)-f(z\,y)) \tag{1}$$ which holds for all $$\,x,y,z>0.\,$$ We want solutions for $$\,f(x)>0.$$ Since this equation holds for all $$\,x,z>0\,$$ both sides depend only on $$\,y.\,$$ call it $$\,g(y).\,$$ Thus if $$\,z=1\,$$ we define $$g(x) := f(1)-f(x). \tag{2}$$ Equations $$(1)$$ and $$(2)$$ imply that $$f(x) - f(x\,y) = g(y)/x. \tag{3}$$ Add equations $$(2)$$ and $$(3)$$ to get $$f(1) - f(x\,y) = g(x) + g(y)/x. \tag{4}$$ Rewrote the left side of equation $$(4)$$ using equation $$(2)$$ to get $$g(x\,y) = g(x) + g(y)/x. \tag{5}$$ Eliminate the denominator in equation $$(5)$$ by defining $$h(x) := x\,g(x). \tag{6}$$ Rewrite equation $$(5)$$ using equation $$(6)$$ to get $$h(x y) = h(x)y + h(y). \tag{7}$$ Define the limit $$c := \lim_{x\to 0} h(x). \tag{8}$$ The limit as $$\,x\to 0\,$$ of equation $$(7)$$ is $$c = c\,y + h(y). \tag{8}$$ Solve equation $$(8)$$ for $$\,h\,$$ to get $$h(y) = c - c\,y. \tag{9}$$ Using equations $$(9)$$, $$(6)$$, and $$(2)$$ to get $$f(x) = f(1) + c - c/x. \tag{10}$$ The original equation in the question was $$f(x)-f(y)=b(f(b\,x)-f(b\,y)). \tag{11}$$ Replacing $$\,y\,$$ by $$\,x\,y\,$$ and $$\,b\,$$ with $$\,z/x\,$$ in this equation is equivalent to our equation $$(1)$$.

EDIT: This version is simpler than my previous one from Jan 22, 2021.

• I understood your method, +1, it is ciertamente correct. Having said that, the other answer was quicker to the same conclusion and also was slightly simpler, which I require since I am using this to answer a question for a beginner user. Thank you very much, also note the comment I made at the end of Professor's answer. Jan 23 at 6:42
• @TeresaLisbon Thanks for your comment! I decided to simplify my answer as much as possible. Maybe you will like it better. Jan 23 at 22:45
• Yes this is much better, thanks Jan 24 at 20:34