Does a solution to this functional equation exist and if so can we construct it?

Question

For $x\geq 0 $ we have $f(x) +xf(1/x) = x/(1+x)$ as well as the conditions $\lim_{x\rightarrow 0} f(x) = 0$ and $\lim_{x\rightarrow \infty} f(x) = 0$. Clearly $f(1) = \frac{1}{4}$. What is the solution, if it exists at all? I have no clue how to tackle this problem. Or are the information not enough to uniquely determine a solution? Thanks.

Answer 1

I played with some intuitive substitutions simplifying the problem stepwise, first define $$f(x)=(1+x)g(x)$$

which leads to

$$g(x)+g(1/x)=\frac{x}{(1+x)^2}$$

Second, substitute

$$g(x)=\frac{1}{2}\frac{x}{(1+x)^2}+h(x)$$ and you get

$$h(x)+h(1/x)=0.\qquad(1)$$

The solution of this is $h(x)=c(x)\ln x$ where $c(x)$ is an arbitrary function which satisfies

$$c(x)-c(1/x)=0.\qquad (2)$$

This leads to

$$f(x)=\frac{x}{2(1+x)}+(1+x)c(x)\ln x.$$

Now let's come back to $c(x)$. (2) is ensured when choosing $$c(x)=d(x)+d(1/x)$$ with an arbitrary function $d(x)$.

Your side conditions now lead to the following remaining question: Find $g(x)$ such that

$$\lim_{x\to 0} (1+x)\left(d(x)+d(1/x)\right)\ln x =0$$ $$\lim_{x\to \infty} (1+x)\left(d(x)+d(1/x)\right)\ln x =-\frac{1}{2}$$ With your side-codition $f(0)=0$ leads to $c=0$, but the second can't be satisfied.

Answer 2

The solution does exist. $$ f(x)=\frac{x}{\left(1+x\right)^2} $$ satisfies the equation: $$ f(x)+x\,f\left(\frac{1}{x}\right)= \frac{x}{\left(x+1\right)^2}+\frac{1}{\left(1+\frac{1}{x}\right)^2}= \frac{x+x^2}{\left(x+1\right)^2}= \frac{x(x+1)}{\left(x+1\right)^2}=\frac{x}{x+1} $$ as required. Unfortunetely, I came up with it by pure guessing, based on the asymptotic behaviour of $f$ near $0$ and $\infty$. I'll try to come up with some more systematic approach.

Answer 3

For any $x > 0$, we have two statements involving $f(x)$ and $f(1/x)$ by applying the equation at $x$ and at $1/x$, and we should make sure that it says the same thing :

At $x$ we get $f(x) + xf(1/x) = x/(1+x)$.
At $1/x$ we get $f(1/x) + (1/x)f(x) = (1/x)/(1+1/x) = 1/(x+1)$, which is equivalent to $xf(1/x) + f(x) = x/(x+1)$ because $ x \neq 0$.
So the two statements are equivalent.

Since your functional equation relates values at $x$ with values at $1/x$, you can pick values of $f$ arbitrarily on $(0;1]$ and this will determine $f$ on $[1;\infty)$ by applying the functional equation.

With your conditions, you need $\lim_{x \to 0} f(0)=0,f(1)=1/4$, and $\lim_{x \to \infty} f(x) = 0$. This means $0 = \lim_{x \to \infty} f(x) = \lim_{x \to 0} f(1/x) = \lim_{x \to 0} 1/(1+x) - f(x)/x = 1 - \lim_{x \to 0} f(x)/x$. So this says that $f'(0)$ exists and is $1$.

Therefore, if you pick any function $f : [0;1] \to \Bbb R$ with $f(0)=0,f'(0)=1,f(1)=1/4$ and extend $f$ on $[1;\infty)$ with $f(x) = x/(1+x)-xf(1/x)$, you obtain a solution to the functional equation.