# Are the only solutions to this implicit functional equation linear?

Let $$f:(0,1] \to (-\infty,0]$$ be a smooth function which is strictly negative on $$(0,1)$$ and satisfies $$f(1)=0$$.

Let $$\epsilon \in (0,1)$$ and let $$x,y:(0,\epsilon) \to (0,1]$$ be smooth functions satisfying $$x(s)y(s)=s$$, $$x(s) < y(s)$$ for every $$s \in (0,\epsilon)$$.

Suppose that $$f \circ x=- y, f \circ y=- x$$, i.e. $$f\big(x(s)\big)=- y(s),f\big(y(s)\big)=- x(s) \tag{1} \label{1}$$ for every $$s \in (0,\epsilon)$$.

Is it true that $$f(x)=x-1$$?

In the case where $$f(x)=x-1$$, equation \eqref{1} implies $$x(s)+y(s)=1$$, which together with $$x(s)y(s)=s$$, implies that $$x(s)=\frac{1-\sqrt{1-4s}}{2}\text,y(s)=\frac{1+\sqrt{1-4s}}{2}\text,$$ so $$x,y$$ are defined on $$\left(0,\frac{1}{4}\right]$$.

The motivation for this question comes from this optimization problem.

Comment:

I specifically assumed a strict inequality $$x. This is important, since otherwise we may get more solutions: Set $$x(s)=y(s)=\sqrt s$$ and $$f(x)=-x$$.

(Although $$f(x)=-x$$ does not satisfy $$f(1)=0$$, so it needs to be modified after $$s=\epsilon$$ in order to satisfy this boundary condition as well).

That's far from true. The following construction gives an uncountable family of counterexamples.

For a given $$\epsilon \in ( 0 , 1 )$$, choose $$\alpha \in \left( 1 , \frac 1 { \sqrt \epsilon } \right)$$ and take any smooth $$x : ( 0 , \epsilon ) \to ( 0 , 1 ]$$ such that:

• $$s \mapsto \frac s { x ( s ) }$$ is strictly decreasing;
• $$c = \lim _ { s \to 0 ^ + } \frac s { x ( s ) } < 1$$;
• $$s \le x ( s ) \le \frac s { \alpha \sqrt \epsilon }$$ for every $$s \in ( 0 , \epsilon )$$.

Define $$y : ( 0 , \epsilon ) \to ( 0 , 1 ]$$ by $$y ( s ) = \frac s { x ( s ) }$$. Then we'll have the following:

• $$x$$ is strictly increasing;
• $$x$$ and $$y$$ are local diffeomorphisms (by the inverse function theorem);
• $$x ( s ) y ( s ) = s$$ for every $$s \in ( 0 , \epsilon )$$;
• there is $$a \in \left( 0 , \frac { \sqrt \epsilon } \alpha \right]$$ such that $$\operatorname { ran } x = ( 0 , a )$$;
• There is $$b \in \left[ \alpha \sqrt \epsilon , c \right)$$ such that $$\operatorname { ran } y = ( b , c )$$;
• $$x ( s ) < y \left( s ' \right)$$ for every $$s , s ' \in ( 0 , \epsilon )$$.

Now define: $$\begin {cases} g : ( 0 , a ) \to ( - \infty , 0 ] \\ g ( t ) = - y \left( x ^ { - 1 } ( t ) \right) \end {cases} \qquad \begin {cases} h : ( b , c ) \to ( - \infty , 0 ] \\ h ( t ) = - x \left( y ^ { - 1 } ( t ) \right) \end {cases}$$ Then $$g$$ and $$h$$ are smooth and strictly increasing functions, for every $$t \in ( 0 , a )$$ we have $$- c < g ( t ) < - b$$, and for every $$t \in ( b , c )$$ we have $$- a < h ( t ) < 0$$. Thus we can mutually extend $$g$$ and $$h$$ to a strictly increasing and smooth $$f : ( 0 , 1 ] \to ( - \infty , 0 ]$$ with $$f ( 1 ) = 0$$, by choosing suitable smooth functions filling $$[ a , b ]$$ and $$[ c , 1 ]$$. Then $$f$$ will satisfy all the desired properties. As we are very much free in choosing $$x$$ and extending $$g$$ and $$h$$, we end up constructing an uncountably infinite family of such functions.

EDIT:

I would like to add that in situations like this, where there is a very large family of smooth functions satisfying certain properties, it may be useful to search for analytic functions in that family. In many cases, that may give a unique solution, which you are seeking. Analytic functions are very rigid and that makes them very rare among smooth functions. For example, the above construction is based on continuing a smooth function in a much free manner. That can't be done for analytic functions, as the analytic continuation is unique. It's interesting that the functions $$x$$, $$y$$ and $$f$$ given by yourself are analytic on their domains. At the moment, I don't know whether that's the only analytic possible case or not.