# Fixed Point of Composition of Inflationary Mappings

## Theorem

Let $\left({S, \preceq}\right)$ be a ordered set.

Let $f, g: S \to S$ be inflationary mappings.

Let $x \in S$.

Then:

$x$ is a fixed point of $f \circ g$
$x$ is a fixed point of both $f$ and $g$.

## Proof

### Necessary Condition

$\Box$

### Sufficient Condition

Let $h = f \circ g$.

Let $x$ be a fixed point of $h$.

Then by the definition of composition:

$f \left({g \left({x}\right)}\right) = x$

Since $f$ is inflationary:

$x \preceq g \left({x}\right)$

Suppose for the sake of contradiction that $x \ne g \left({x}\right)$.

Then $x \prec g \left({x}\right)$.

Since $f$ is also inflationary:

$g \left({x}\right) \preceq f \left({g \left({x}\right)}\right)$

Thus by Extended Transitivity:

$x \prec f \left({g \left({x}\right)}\right)$

But this contradicts the assumption that $x$ is a fixed point of $f \circ g$.

Therefore, $x = g \left({x}\right)$, and $x$ is a fixed point of $g$.

Suppose for the sake of contradiction that $f \left({x}\right) \ne x$.

Then $x \prec f \left({x}\right)$.

As we have shown that $x = g \left({x}\right)$, it follows that:

$x \prec f \left({g \left({x}\right)}\right)$

But this contradicts assumption that $x$ is a fixed point of $f \circ g$.

Hence, $x$ is also a fixed point of $f$.

$\blacksquare$