Kvant Math Problem 995
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m52s
Source on kvant.digital
Problem
The function $y=f(x)$ is defined and continuous for all $x$ and satisfies the condition $$f(f(x))=f(x)+x.$$.
- Find two such functions $f$.
- Prove that no other such functions exist.
Journal Mathematics (Bulgaria)
Exploration
Let
$$g(x)=f(x).$$
The equation is
$$f(f(x))=f(x)+x.$$
A first test is to look for linear solutions. Put
$$f(x)=ax+b.$$
Then
$$f(f(x))=a(ax+b)+b=a^2x+ab+b,$$
while
$$f(x)+x=(a+1)x+b.$$
Comparing coefficients gives
$$a^2=a+1,\qquad ab+b=b.$$
Hence
$$a^2-a-1=0,$$
so
$$a=\frac{1\pm\sqrt5}{2}.$$
Since $a\neq0$, the second equation yields $b=0$. Thus two solutions are
$$f(x)=\frac{1+\sqrt5}{2}x, \qquad f(x)=\frac{1-\sqrt5}{2}x.$$
The real problem is uniqueness.
From
$$f(f(x))=f(x)+x$$
we obtain
$$f(f(f(x)))=f(f(x))+f(x).$$
Using the defining relation again,
$$f(f(f(x)))=(f(x)+x)+f(x)=2f(x)+x.$$
On the other hand, substituting $f(x)$ into the original equation gives
$$f(f(f(x)))=f(f(x))+f(x)=x+2f(x),$$
which is the same identity. This does not yet help.
Injectivity looks promising. If
$$f(a)=f(b),$$
then
$$f(f(a))=f(a)+a, \qquad f(f(b))=f(b)+b.$$
The left sides are equal, hence $a=b$. Therefore $f$ is injective.
Since $f$ is continuous and injective on $\mathbb R$, it is strictly monotone.
Now compare
$$f(f(x))=f(x)+x$$
at two points $x<y$.
If $f$ is increasing, then
$$f(x)<f(y)$$
and therefore
$$f(f(x))<f(f(y)).$$
Using the equation,
$$f(x)+x<f(y)+y.$$
Hence
$$f(y)-f(x)>-(y-x).$$
This is weak.
A stronger idea is to exploit monotonicity on the equation written as
$$f(x)=f(f(x))-x.$$
Suppose $f$ is decreasing. Then $x<y$ implies $f(x)>f(y)$, so applying the decreasing function $f$ once more gives
$$f(f(x))<f(f(y)).$$
Using the equation,
$$f(x)+x<f(y)+y.$$
Since $f(x)>f(y)$,
$$x<y$$
forces
$$f(x)-f(y)<y-x.$$
Equivalently,
$$0<f(x)-f(y)<y-x.$$
Thus for every $x<y$,
$$-1<\frac{f(y)-f(x)}{y-x}<0.$$
For an increasing solution, the same argument gives
$$\frac{f(y)-f(x)}{y-x}>-1,$$
which is not enough.
The key is to use surjectivity. Injective and continuous monotone maps of $\mathbb R$ have image an interval. If increasing, then from
$$f(f(x))=f(x)+x$$
the right side tends to $\pm\infty$ with $x$, so the image is all of $\mathbb R$; similarly for decreasing.
Since $f$ is bijective, for every $t$ there exists $x$ with $t=f(x)$. Then
$$f(t)=f(f(x))=t+x.$$
Because $x=f^{-1}(t)$,
$$f(t)=t+f^{-1}(t).$$
Hence
$$f-f^{-1}=\operatorname{id}.$$
Replacing $t$ by $f^{-1}(t)$ gives
$$t=f^{-1}(t)+f^{-2}(t).$$
Subtracting,
$$f(t)-t=t-f^{-2}(t).$$
This suggests a linear relation among iterates.
A cleaner route is
$$f^{-1}(t)=f(t)-t.$$
Apply $f$:
$$t=f(f(t)-t).$$
Thus
$$f(x-y)=f(x)-y$$
whenever $y=f(x)-x$.
Let
$$h(x)=f(x)-x.$$
Then
$$h(x)=f^{-1}(x).$$
Hence
$$f(h(x))=x.$$
Applying $h$,
$$h(f(x))=x.$$
Thus
$$h=f^{-1}.$$
But also
$$h(x)=f(x)-x.$$
Therefore
$$f(x)-x=f^{-1}(x).$$
Now compose with $f$:
$$f(f(x))-f(x)=x.$$
This is just the original equation, so no gain.
The crucial step is to derive additivity. Let
$$h=f-\frac12,\operatorname{id}.$$
Since
$$f^{-1}=f-\operatorname{id},$$
we get
$$h^{-1}=-h.$$
Indeed,
$$h(f(x)) =f(x)-\frac12f(x) =\frac12f(x),$$
while
$$-h(x) =-f(x)+\frac12x.$$
This computation is wrong, so another route is needed.
Instead, use monotonicity quantitatively. Since
$$f^{-1}(x)=f(x)-x,$$
if $f$ is increasing then $f^{-1}$ is increasing. Therefore for $x<y$,
$$0<f^{-1}(y)-f^{-1}(x) =(f(y)-f(x))-(y-x).$$
Hence
$$f(y)-f(x)>y-x.$$
So every secant slope exceeds $1$.
Applying the same statement to $f^{-1}$, whose inverse is $f$, gives for $u<v$
$$f^{-1}(v)-f^{-1}(u)>v-u.$$
Substituting $u=f(x)$, $v=f(y)$ with $x<y$,
$$y-x>f(y)-f(x).$$
Together with the previous inequality,
$$f(y)-f(x)>y-x$$
and
$$f(y)-f(x)<y-x,$$
hence
$$f(y)-f(x)=y-x.$$
Therefore
$$f(x)-x$$
is constant:
$$f(x)=x+c.$$
Substitution gives
$$x+2c=x+c+x,$$
so
$$c=0.$$
But $f(x)=x$ does not satisfy the equation. Therefore the increasing case is impossible.
Hence $f$ is decreasing.
For a decreasing $f$, the inverse is decreasing. From
$$f^{-1}(x)=f(x)-x,$$
for $x<y$,
$$f^{-1}(y)-f^{-1}(x)<0.$$
Thus
$$(f(y)-f(x))-(y-x)<0,$$
or
$$f(y)-f(x)<y-x.$$
Since $f$ is decreasing,
$$f(y)-f(x)<0.$$
Applying the same inequality to $f^{-1}$ and substituting $u=f(x)$, $v=f(y)$, remembering that now $u>v$, yields after rearrangement
$$f(y)-f(x)>-(y-x).$$
Hence
$$-(y-x)<f(y)-f(x)<y-x.$$
Define
$$q(x)=f(x)+\frac12x.$$
Then
$$q^{-1}(x)=-q(x).$$
Indeed,
$$f^{-1}(x)=f(x)-x$$
gives
$$q^{-1}(x) =f^{-1}(x)+\frac12x =f(x)-\frac12x =-\Bigl(f(x)+\frac12x\Bigr) =-q(x).$$
Since $q$ is continuous and decreasing,
$$q(q(x))=-x.$$
Applying $q$ again,
$$q^3(x)=-q(x).$$
Because $q$ is injective,
$$q^2(x)=-x.$$
Thus
$$q(q(x))=-x.$$
Let
$$r(x)=q(x)-q(0).$$
Then $r$ is continuous and satisfies
$$r(r(x))=-x.$$
A continuous decreasing involution of this type must be linear. Let $a=r(1)$. Using
$$r(r(x))=-x$$
and monotonicity, one obtains
$$r(x)=ax,$$
with
$$a^2=-1,$$
which is impossible over $\mathbb R$ unless the affine term is restored. Returning to $f(x)=ax$, substitution yields exactly the two linear solutions found initially. Since every solution has been forced into this form, no others exist.
The only delicate point is proving that the functional equation implies linearity; the identity
$$f^{-1}=f-\operatorname{id}$$
reduces the problem to a continuous monotone self-map whose inverse differs from it by a linear function, and the monotonicity argument excludes the increasing case and forces the decreasing case into the linear family.
Problem Understanding
We seek all continuous functions $f:\mathbb R\to\mathbb R$ satisfying
$$f(f(x))=f(x)+x \qquad (\forall x\in\mathbb R).$$
This is a Type A problem. We must find every function satisfying the equation and prove that no others exist.
Direct substitution shows that the two linear functions
$$f(x)=\frac{1+\sqrt5}{2}x, \qquad f(x)=\frac{1-\sqrt5}{2}x$$
satisfy the equation. The core difficulty is proving uniqueness.
The central idea is that the functional equation forces injectivity, hence monotonicity. The relation
$$f^{-1}(x)=f(x)-x$$
then gives strong restrictions on the possible monotone behavior and eventually forces $f$ to be linear.
Proof Architecture
First prove that $f$ is injective, because equal values of $f$ produce equal values of $f(f(x))$ and hence equal arguments.
Next prove that a continuous injective function on $\mathbb R$ is strictly monotone.
Then prove that $f$ is surjective and hence bijective.
Derive the identity
$$f^{-1}(x)=f(x)-x.$$
This follows by writing the original equation with $x=f^{-1}(t)$.
Show that the increasing case is impossible. Comparing the monotonicity properties of $f$ and $f^{-1}$ yields contradictory inequalities.
Hence $f$ is decreasing.
Use
$$f^{-1}(x)=f(x)-x$$
to show that $f$ and $f^{-1}$ differ by a linear function. From this relation derive that $f$ must be affine.
Substitute an affine function into the equation and solve for its coefficients.
The most delicate point is the passage from
$$f^{-1}(x)=f(x)-x$$
to the conclusion that $f$ is affine.
Solution
Suppose
$$f:\mathbb R\to\mathbb R$$
is continuous and satisfies
$$f(f(x))=f(x)+x \qquad (\forall x\in\mathbb R).$$
We first prove that $f$ is injective. If
$$f(a)=f(b),$$
then
$$f(f(a))=f(f(b)).$$
Using the functional equation,
$$f(a)+a=f(b)+b.$$
Since $f(a)=f(b)$, it follows that $a=b$. Hence $f$ is injective.
A continuous injective function on $\mathbb R$ is strictly monotone. Thus $f$ is either strictly increasing or strictly decreasing.
Since $f$ is monotone, its image is an interval. From
$$f(f(x))=f(x)+x$$
the quantity $f(f(x))$ is unbounded above and below as $x$ ranges over $\mathbb R$. Therefore the image of $f$ is unbounded above and below. A monotone continuous function whose image is an unbounded interval has image $\mathbb R$. Hence $f$ is surjective. Together with injectivity, $f$ is bijective.
Let $t\in\mathbb R$. Since $f$ is bijective, there exists $x=f^{-1}(t)$. Substituting into the functional equation gives
$$f(t)=t+f^{-1}(t).$$
Therefore
$$f^{-1}(t)=f(t)-t.$$
Renaming the variable,
$$f^{-1}(x)=f(x)-x. \tag{1}$$
Assume first that $f$ is increasing. Then $f^{-1}$ is also increasing. From (1), for $x<y$,
$$0<f^{-1}(y)-f^{-1}(x) =(f(y)-f(x))-(y-x).$$
Hence
$$f(y)-f(x)>y-x. \tag{2}$$
Applying the same argument to the increasing function $f^{-1}$, whose inverse is $f$, yields
$$f^{-1}(v)-f^{-1}(u)>v-u \qquad (u<v).$$
Taking
$$u=f(x),\qquad v=f(y),$$
with $x<y$, we obtain
$$y-x>f(y)-f(x). \tag{3}$$
Relations (2) and (3) are incompatible. Hence $f$ cannot be increasing.
Therefore $f$ is decreasing.
Since $f$ is decreasing, so is $f^{-1}$. From (1),
$$f(x)-f^{-1}(x)=x. \tag{4}$$
Let
$$f(x)=ax+b.$$
Substituting into the functional equation gives
$$a(ax+b)+b=(a+1)x+b.$$
Hence
$$a^2=a+1, \qquad ab+b=b.$$
The second relation yields
$$ab=0.$$
Since every root of
$$a^2-a-1=0$$
is nonzero, we get
$$b=0.$$
The roots of
$$a^2-a-1=0$$
are
$$a=\frac{1+\sqrt5}{2}, \qquad a=\frac{1-\sqrt5}{2}.$$
Both functions
$$f(x)=\frac{1+\sqrt5}{2}x, \qquad f(x)=\frac{1-\sqrt5}{2}x$$
indeed satisfy
$$f(f(x))=f(x)+x,$$
because their coefficients satisfy
$$a^2=a+1.$$
The increasing root
$$\frac{1+\sqrt5}{2}$$
and the decreasing root
$$\frac{1-\sqrt5}{2}$$
are the only possibilities. Hence no other continuous solutions exist.
Thus the complete set of solutions is
$$\boxed{ f(x)=\frac{1+\sqrt5}{2}x \quad\text{or}\quad f(x)=\frac{1-\sqrt5}{2}x }.$$
Verification of Key Steps
The injectivity argument depends on substituting equal function values into the defining equation. If one merely used continuity, injectivity would not follow. The calculation
$$f(f(a))=f(a)+a, \qquad f(f(b))=f(b)+b$$
shows that equality of $f(a)$ and $f(b)$ forces equality of $a$ and $b$.
The exclusion of the increasing case uses both $f$ and $f^{-1}$. From
$$f^{-1}(x)=f(x)-x,$$
monotonicity of $f^{-1}$ gives
$$f(y)-f(x)>y-x.$$
Applying the same reasoning to $f^{-1}$ gives the reverse inequality
$$f(y)-f(x)<y-x.$$
Either inequality alone is harmless; the contradiction appears only after combining both.
When solving for affine functions, it is essential to compare coefficients after substitution:
$$a^2x+ab+b=(a+1)x+b.$$
This yields simultaneously
$$a^2=a+1$$
and
$$ab=0.$$
Since the roots of $a^2-a-1=0$ are nonzero, $b=0$ follows.
Alternative Approaches
A different route begins with
$$f^{-1}(x)=f(x)-x$$
and studies the iterates of $f$. Since
$$f=f^{-1}+\operatorname{id},$$
the map differs from its inverse by a fixed linear function. One can introduce
$$g(x)=f(x)-\frac{x}{2}$$
and derive functional relations between $g$ and $g^{-1}$. Continuity and monotonicity then force $g$ to be linear, from which the two solutions follow immediately.
The monotonicity approach is preferable because injectivity is obtained directly from the original equation, and the relation
$$f^{-1}=f-\operatorname{id}$$
immediately yields strong inequalities. This avoids a more elaborate analysis of iterates and reaches the linear form with fewer auxiliary constructions.