Two real functions with a given composition

Question

We are looking for some (all) functions $g:\mathbb{R}\to \mathbb{R}$ and $f:\mathbb{R}^2\to \mathbb{R}$ with $\min\{x,y\}<f(x,y)<\max\{x,y\}$ (if $x\neq y$) and satisfying the functional equation $$g(f(x,y))=x^2+xy+y^2.$$

Note. It is easy to see that for the real function $g(t)=3t^2$ there exists $f(x,y)$ such that $|f(x,y)|=\sqrt{\frac{x^2+xy+y^2}{3}}$, and all the mentioned conditions hold.

Any other solutions?

Answer 1

Your example is already flawed because the square root does not connect ℝ^2 with ℝ but with the complexes on the right hand side of the equation. The absolute value has to be taken for a match on this side too. Think on x<0 but |x|>y and so on. Everything has to be real on the right hand-side. And is valid despite taking the absolute value is idempotent and a not bijective.

A good idea is matrices:

g(f(x,y))=(x,y)((1,0.5),(0.5,1))(x,y)T

this is indeed a function form the ℝ^2 in ℝ. And it is a representation of the composite function given in the question. T transpose the raw pair (x,y).

A pair of constant factors as in Your example to the rest.

g(t)=2 t and f(x,y)=(x,y)((1,0.5),(0.5,1))(x,y)T/2.

Any other pairing will do too:

g(t)= t/Sin(t) and f(x,y)=Sin((x,y)((1,0.5),(0.5,1))(x,y)T/2)*(x,y)((1,0.5),(0.5,1))(x,y)T/2

This is stating that a composite function hide a quotient of everything mathematical as far as the denominator and the nominator are distributed in the right fashion into the function applied first and the function applied second. The second example indicated that the outer functional or polynomial dependence must be the same. So the first applied function must be a composite function too and the composition has to be inverted be g(ℝ)->ℝ. So f=g^-1(h(x,y)).

g(ℝ)->ℝ is not to completely free since the space of invertible is not large compared to all functions possible.

So the question rewrites to: g(g^-1(h(x,y))=h(x,y).

What are the conditions good for?

min{,}<(,)<max{,} (if ≠)

min and max problems always require a case analysis.

x>y: min(x,y)=x: y<f(x,y) and max(x,y)=x>f(x,y), so y<f(x,y)<x

x<y: mix(x,y)=y: x<f(x,y) and max(x,y)=y>f(x,y), so x<f(x,y)<y

This shows the range in which the function f can be.

max(x,y)>=f(x,y) with the given f:

f(x,y)>=min(x,y) with the given f:

And the overlap logical AND:

Each as two dimensional regions.

So this abstract cases can be rather difficult in the region they are valid.

As the first plot {min,max}:ℝ2→ℝ shows this are always divergent function classes.

f is in this question bilinear form with a determinant that has a positive determinant so it is really subjective on ℝ2→ℝ. It leads to this characteristic two ellipsis and is not valid around a overlap in the vicinity of zero of the two ellipsis. So this bilinear form is affine and the coordinate system is in need of the affine move and a main axis rotation to get a bilinear form in main axis representation. Bilinear forms are different from polynomials in general with special attributes and the quartic simplicity. They are polynomials with two variables and of order two.

In this region in inequality chain is really narrow: