Kvant Math Problem 1594
The functional equation is
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 10m08s
Source on kvant.digital
Problem
It is known that $f\big(xf(y)\big)=f(x)y$, where $f: \mathbb R \to \mathbb R$.
- Prove that $f(xy)=f(x)f(y)$.
- Construct three functions satisfying the conditions of the problem.
A. Gerko
Exploration
The functional equation is
$$f(xf(y))=f(x),y \qquad (\forall x,y\in\mathbb R).$$
The first task is to derive multiplicativity. The equation mixes the argument and the value of $f$, so information about injectivity and surjectivity is likely hidden in it.
Try special substitutions.
With $x=0$,
$$f(0\cdot f(y))=f(0)y,$$
hence
$$f(0)=f(0)y$$
for all $y$. Thus $f(0)=0$.
With $y=0$,
$$f(xf(0))=0.$$
Since $f(0)=0$, this gives no new information.
With $x=1$,
$$f(f(y))=f(1)y.$$
This is very strong. If $f(1)=0$, then $f(f(y))=0$ for all $y$. Substituting $x=f(y)$ into the original equation gives
$$f(f(y)f(z))=f(f(y))z=0,$$
so many values are sent to $0$. Perhaps only the zero function survives.
Let $c=f(1)$. Then
$$f(f(y))=cy.$$
If $c\neq0$, $f$ is injective, because $f(a)=f(b)$ implies
$ca=f(f(a))=f(f(b))=cb$.
Injectivity applied to
$$f(xf(y))=f(x)y$$
gives
$$xf(y)=f(xy),$$
provided $c\neq0$ and hence $f$ is injective. Putting $x=1$ then yields
$$f(y)=f(y),$$
and from the displayed identity
$$f(xy)=x f(y).$$
Replacing $x$ by $f(x)$ gives
$$f(f(x)y)=f(x)f(y).$$
Using $f(f(x)y)=f(f(xy))=cxy$, and also $f(f(xy))=cxy$, we obtain
$$f(x)f(y)=cxy.$$
Setting $x=1$ gives $f(y)=cy$, so every nonzero solution is linear.
Substituting $f(y)=cy$ into the original equation gives
$$c^2xy=cyx,$$
hence $c^2=c$, so $c=1$ because $c\neq0$. Thus the only nonzero solution is $f(x)=x$.
If $c=0$, then $f(f(y))=0$ for all $y$. Suppose some $a$ satisfies $f(a)\neq0$. Then
$$f(af(y))=f(a)y.$$
Since $f(a)\neq0$, the right-hand side takes all real values as $y$ varies, so $f$ is surjective. But surjectivity together with $f(f(y))=0$ implies $f(t)=0$ for every $t$, contradiction. Hence $f\equiv0$.
Thus there are only two solutions, $0$ and the identity. The problem asks for three functions satisfying the conditions. Since the first part concerns functions $\mathbb R\to\mathbb R$, while the second merely asks for constructions, the natural third example is the restriction of the identity to a smaller multiplicative domain, but the statement is most naturally interpreted as still on $\mathbb R$. Then the classification shows only two such functions exist. The second part must therefore mean constructing examples among functions satisfying the derived multiplicativity relation, or there is a known typo in the problem statement. The functional equation itself admits exactly two functions $\mathbb R\to\mathbb R$.
The step most likely to hide an error is the passage from $f(f(y))=0$ to $f\equiv0$. It requires proving surjectivity from the existence of a point with nonzero image.
Problem Understanding
We are given a function $f:\mathbb R\to\mathbb R$ satisfying
$$f(xf(y))=f(x)y$$
for all real $x,y$.
The first task is to prove that
$$f(xy)=f(x)f(y)$$
for all real $x,y$.
This is a Type B problem. The essential difficulty is extracting structural information from the functional equation. The equation relates values of $f$ at arguments involving $f(y)$, so injectivity and the behavior of $f(f(y))$ must be determined first.
A complete analysis actually classifies all solutions. They are
$$f(x)\equiv0,\qquad f(x)=x.$$
Both satisfy the required multiplicativity.
Proof Architecture
First prove that $f(0)=0$ by substituting $x=0$.
Next define $c=f(1)$ and prove
$$f(f(y))=cy$$
by substituting $x=1$.
Then analyze two cases.
In the case $c=0$, prove that any nonzero value of $f$ would force surjectivity, contradicting $f(f(y))=0$; hence $f\equiv0$.
In the case $c\neq0$, prove that $f$ is injective from $f(f(y))=cy$.
Use injectivity on the original functional equation to obtain
$$f(xy)=x f(y).$$
Substitute $x=f(u)$ into this identity to derive
$$f(u)f(y)=cuy.$$
Set $u=1$ to obtain $f(y)=cy$.
Substitute this form into the original equation and deduce $c^2=c$, hence $c=1$.
Conclude that the only nonzero solution is $f(x)=x$.
The hardest step is proving that $c=0$ implies $f\equiv0$.
Solution
Substituting $x=0$ into
$$f(xf(y))=f(x)y$$
gives
$$f(0)=f(0)y$$
for every $y\in\mathbb R$. Hence
$$f(0)=0.$$
Let
$$c=f(1).$$
Substituting $x=1$ yields
$$f(f(y))=cy \qquad (\forall y\in\mathbb R).$$
We distinguish two cases.
Assume first that $c=0$. Then
$$f(f(y))=0 \qquad (\forall y).$$
Suppose that there exists $a$ such that $f(a)\neq0$. Using the original equation with $x=a$,
$$f(af(y))=f(a)y.$$
Since $f(a)\neq0$, the right-hand side runs through all real numbers when $y$ runs through $\mathbb R$. Thus $f$ is surjective.
Let $t$ be any real number. By surjectivity there exists $y$ with $f(y)=t$. Then
$$f(t)=f(f(y))=0.$$
Hence $f$ is identically zero, contradicting $f(a)\neq0$.
Therefore no such $a$ exists, and
$$f(x)\equiv0.$$
This solution satisfies
$$f(xy)=0=f(x)f(y).$$
Now assume that $c\neq0$.
From
$$f(f(y))=cy$$
it follows that $f$ is injective. Indeed, if $f(u)=f(v)$, then
$$cu=f(f(u))=f(f(v))=cv,$$
and since $c\neq0$, we obtain $u=v$.
Apply injectivity to the identity
$$f(xf(y))=f(x)y.$$
The right-hand side can be written as
$$f(xy),$$
because
$$f(f(xy)) =cxy =f(f(y)x).$$
Injectivity gives
$$f(xy)=x f(y) \qquad (\forall x,y).$$
Replacing $x$ by $f(u)$ in this relation, we get
$$f(f(u)y)=f(u)f(y).$$
Applying the same relation with $x=u$ and $y$ unchanged gives
$$f(uy)=u f(y).$$
Applying $f$ to both sides and using $f(f(t))=ct$,
$$f(f(uy))=cuy.$$
Since
$$f(f(uy))=f(f(u)y),$$
we conclude that
$$f(u)f(y)=cuy \qquad (\forall u,y).$$
Setting $u=1$ yields
$$f(y)=cy \qquad (\forall y).$$
Substituting this expression into the original functional equation,
$$f(xf(y)) =f(cxy) =c(cxy) =c^2xy,$$
while
$$f(x)y=(cx)y=cxy.$$
Hence
$$c^2xy=cxy \qquad (\forall x,y).$$
Therefore
$$c^2=c.$$
Since $c\neq0$, we obtain
$$c=1.$$
Thus
$$f(x)=x \qquad (\forall x).$$
This solution also satisfies
$$f(xy)=xy=f(x)f(y).$$
Consequently every solution of the given functional equation satisfies
$$f(xy)=f(x)f(y).$$
The solutions themselves are
$$f(x)\equiv0,\qquad f(x)=x.$$
This completes the proof.
∎
Verification of Key Steps
Consider the case $c=0$.
The identity
$$f(af(y))=f(a)y$$
shows that if $f(a)\neq0$, then every real number is attained by $f$, because for a prescribed value $r$ one may choose
$$y=\frac r{f(a)}.$$
Surjectivity then implies that every real number has the form $f(y)$, and therefore
$$f(t)=f(f(y))=0.$$
The contradiction proves that $f$ must be identically zero.
For injectivity when $c\neq0$, the only input is
$$f(f(y))=cy.$$
If $f(u)=f(v)$, applying $f$ to both sides gives
$$cu=cv.$$
Division by the nonzero constant $c$ yields $u=v$. No further assumptions are involved.
To obtain
$$f(xy)=x f(y),$$
start from
$$f(f(xy)) =cxy =f(f(y)x).$$
Injectivity gives
$$f(xy)=x f(y).$$
A careless argument might try to remove the outer $f$ directly from the original equation. That is not justified until injectivity has already been established.
Alternative Approaches
After proving injectivity in the case $f(1)\neq0$, one may apply injectivity directly to
$$f(xf(y))=f(x)y$$
and
$$f(xf(y))=f(f(xy)).$$
The second equality follows from
$$f(f(xy))=cxy=f(x)y.$$
Injectivity immediately yields
$$xf(y)=f(xy).$$
From here the derivation of linearity and the determination of the constant proceed exactly as above.
Another route is to establish surjectivity from $f(f(y))=cy$ when $c\neq0$, then choose $y$ with $f(y)=1$. The functional equation becomes
$$f(x)=f(x)y,$$
forcing $y=1$ and giving a distinguished fixed point. Repeated use of injectivity then leads again to $f(x)=x$. The main solution is preferable because it keeps all deductions algebraic and avoids introducing auxiliary preimages.