Name for functions that are anti-symmetric about $y=x$

Question

Even functions are functions that are symmetric about the $Y$-axis, and odd functions are functions that are symmetric about the origin. Functions that are symmetric about $y=x$ ($y=f(x)$ implies $x=f(y)$) are involutions, i.e. functions that are their own inverse.

Is there a special name for functions that are anti-symmetric about $y=x$? In other words, is there a name for the property: If $y=f(x)$ and $x=f(y)$ then $x=y$? The word anti-involution seems to be in use, but according to Wikipedia it has a rather technical definition in terms of antihomomorphisms and doesn't seem to be what I'm looking for.

Answer 1

Any monotonic increasing function works. If $f(x)<x$ then $f(f(x))\leq f(x)<x,$ so $f(f(x))\neq x.$ Similarly, $f(x)>x,$ then $f(f(x))\geq f(x)>x,$ so again $f(f(x))\neq x.$ You don't need $f$ to be strictly monotonic.

Also, any $f$ such that $f(x)\geq x$ for all $x.$ If $f(x)\neq x,$ then $f(x)>x$ and $f(f(x))\geq f(x)>x.$ This includes functions like $f(x)=x+x^2,$ or more generally $f(x)=x+g(x)^2,$ for any function $g,$ which are not strictly increasing.

Similarly, if $f(x)\leq x$ for all $x,$ then when $f(x)\neq x$ we get $f(x)<x$ and thus $f(f(x))\leq f(x)<x.$ So this includes functions like $f(x)=x-g(x)^2.$

$x+g(x)^2$ and $x-g(x)^2$ allows solutions to $g(x)=0,$ so we can get functions $f$ with arbitrary sets of solutions to $f(x)=0.$ Given any closed $C\subseteq\mathbb R,$ we can find continuous $g$ with $C=\{x\mid g(x)=0\}$ and then $C=\{x\mid f(x)=x\}.$

There are a lot of such functions.

Given a general $f,$ if $h(x)=f(x)-x,$ your condition $f(f(x))=x$ becomes $h(x+h(x))+h(x)=0$ so you want $$h(x+h(x))+h(x)=0\implies h(x)=0.\tag1$$

It is worth considering what $(1)$ means if $f$ is monotonic increasing.

If $h$ is differentiable, monotonicity of $f$ means $h'(x)\neq -2$ for all $x.$ Then $h(x+h(x))=h(x)+h'(c)h(x)$ for some $c$ in $(x,x+h(x)).$ So $0=h(x+h(x))+h(x)=(2+h'(c))h(x).$ But $2+h'(c)\neq 0,$ so this means $h(x)=0.$

So we can apparently use any $h(x)$ differentiable such that for all $x,$ $h'(x)\neq -2.$ So this means and differentiable $f(x)$ such that $f'(x)\neq -1.$

If $f$ is differentiable, and $f'(x)\neq -1$ for all $x,$ we have $f(f(x))=x$ if and only if $f(x)=x.$

Proven directly: If $u=f(x)$ and $u\neq x,$ then $f(u)=x$ and $u-x=f(x)-f(u)=f'(c)(x-u)$ for some $c$ between $u$ and $x.$ But $f'(c)\neq -1,$ this isn't possible.

So this includes a lot of functions which decrease "too quickly," like $f(x)=-2x.$

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Darboux's Theorem says that any function which is the derivative of another satisfies the intermediate value property, so if $f$ is differentiable and $f'(x)\neq -1$ for all $x,$ then either $f'(x)<-1$ for all $x$ or $f'(x)>-1$ for all $x.$

If course, there are $f$ with some $f'(x)=-1$ which satisfy this condition. For example, when $f(x)=-(x+x^3).$ Then $f'(0)=-1,$ but $f(f(x))=x$ still implies $f(x)=x.$ This is because while $f'(0)=-1,$ there is no $u\neq v$ with $f(u)-f(v)=v-u,$ because the function $f$ crosses the tangent at $x=0.$

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Of course, our mean value condition on $u,v$ excluded $f(u)+u=f(v)+v$ when $u\neq v,$ which is stronger than excluding $f(u)=v, f(v)=u.$

For example, $f(x)=x^2-x$ has $f(f(x))-x = x^4-2x^3,$ so the roots of $f(f(x))=x$ are $x=0,2$ but $f(0)=0, f(2)=2.$ So $f$ has your property, but $f(x)+x = f(-x)+(-x)=x^2,$ so $f$ has infinitely many pairs $u\neq v$ with $f(u)+u=f(v)+v$ but no such pair with $f(u)=v$ and $f(v)=u.$

Answer 2

The function $f^{-1}(x)$ and f(x) are symmetric to the line y=x and if $y=f(x) -> x=f^{-1}(y)$ is that what you mean?Example $f(x)=x^2, x>=0 and f(x)=\sqrt{x}$ are symmetric to y=x