Kvant Math Problem 2859

Verified: yes
Verdicts: PASS + PASS
Solve time: 11m34s
Source on kvant.digital

Problem

Determine whether there exists a function $f: \mathbb{R} \to \mathbb{R}$ such that for all $x \in \mathbb{R}$ and $y \in \mathbb{R}$ satisfying $x\lt y$, the inequality $$f(x)\gt f(f(y)).$$ holds.

L. Shatunov

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Exploration

Assume that a function $f:\mathbb R\to\mathbb R$ satisfies

$$x<y\implies f(x)>f(f(y)).$$

The previous argument relied on passing to the limit $y\to x^+$, which is unavailable because no regularity assumptions are given. A different route is needed.

The inequality compares values of $f$ at arbitrary points with values of the second iterate. Since it holds for every smaller $x$, fixing $y$ shows that $f(f(y))$ is a strict lower bound for the set

$${f(x):x<y}.$$

This suggests examining the image of $f$. If some value belongs to the image, applying the defining inequality with a suitable choice of variables may force a contradiction inside the image itself.

The key idea is to choose $x=f(y)$ whenever possible and compare $f(f(y))$ with itself.

Problem Understanding

We seek to determine whether there exists a function $f:\mathbb R\to\mathbb R$ such that

$$f(x)>f(f(y))$$

whenever $x<y$.

The problem is of existence type. A complete solution requires either constructing such a function or deriving a contradiction from the assumption that one exists.

The crucial difficulty is that the inequality involves the second iterate of $f$, while no continuity or monotonicity assumptions are available.

Proof Architecture

Assume that a function satisfying the condition exists.

For an arbitrary real number $y$, compare the numbers $f(y)$ and $y$.

If $f(y)<y$, then the defining condition applied to the pair $(f(y),y)$ gives

$$f(f(y))>f(f(y)),$$

which is impossible.

Hence every $y$ satisfies

$$f(y)\ge y.$$

Since $f(y)\ge y$, applying the defining condition to the pair $(y,f(y))$ whenever $f(y)>y$ yields

$$f(y)>f(f(f(y))).$$

Replacing $y$ by $f(y)$ in the previously established inequality $f(t)\ge t$ gives

$$f(f(y))\ge f(y).$$

Thus $f(y)$ cannot exceed $y$, and consequently $f(y)=y$ for every $y$.

The identity function plainly does not satisfy the original inequality, producing the desired contradiction.

Solution

Suppose that a function $f:\mathbb R\to\mathbb R$ satisfies

$$x<y\implies f(x)>f(f(y)).$$

Fix an arbitrary real number $y$.

Assume first that $f(y)<y$. Since the hypothesis applies to every pair with smaller first coordinate, we may substitute

$$x=f(y),\qquad y=y.$$

Because $f(y)<y$, the condition gives

$$f(f(y))>f(f(y)),$$

which is absurd.

Hence

$$f(y)\ge y$$

for every real number $y$.

Now suppose that for some $y$ we had $f(y)>y$. Then the pair

$$x=y,\qquad y=f(y)$$

satisfies $x<y$, so the defining condition yields

$$f(y)>f(f(f(y))).$$

On the other hand, the inequality already proved, applied to the number $f(y)$, gives

$$f(f(y))\ge f(y).$$

Applying the same statement once more to the number $f(f(y))$ gives

$$f(f(f(y)))\ge f(f(y)).$$

Combining these inequalities,

$$f(f(f(y)))\ge f(f(y))\ge f(y),$$

which contradicts

$$f(y)>f(f(f(y))).$$

Thus no such $y$ can exist, and therefore

$$f(y)\le y$$

for every real number $y$.

Together with the previously established inequality $f(y)\ge y$, this implies

$$f(y)=y$$

for all $y\in\mathbb R$.

Hence $f$ must be the identity function.

Substituting $f(x)=x$ into the original condition gives

$$x<y\implies x>y,$$

which is impossible.

Accordingly, no function $f:\mathbb R\to\mathbb R$ satisfying

$$x<y\implies f(x)>f(f(y))$$

exists.

$$\boxed{\text{No such function exists}.}$$

Verification of Key Steps

No continuity argument is used. Every application of the hypothesis involves a pair of numbers whose order is explicitly known.

The deduction $f(y)\ge y$ follows from the contradiction obtained by choosing $x=f(y)$ when $f(y)<y$.

The opposite inequality is obtained by assuming $f(y)>y$, applying the hypothesis to the pair $(y,f(y))$, and comparing this with the consequences of $f(t)\ge t$ for $t=f(y)$ and $t=f(f(y))$.

These two inequalities force $f(y)=y$ for every $y$, and the identity function violates the original condition.

Alternative Approaches

Another route starts from the observation that $f(y)<y$ is impossible. Once $f(y)\ge y$ is known for all $y$, the iterates satisfy

$$y\le f(y)\le f(f(y))\le \cdots.$$

If some strict inequality $y<f(y)$ occurred, then applying the original condition to the pair $(y,f(y))$ would give

$$f(y)>f(f(f(y))),$$

while the monotonic chain of iterates implies

$$f(f(f(y)))\ge f(y),$$

which is contradictory.

This again forces $f(y)=y$ for all $y$, leading to the same contradiction.