IMO 1972 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 14m14s

Problem

Let $f$ and $g$ be real-valued functions defined for all real values of $x$ and $y$, and satisfying the equation $$ f(x + y) + f(x - y) = 2f(x)g(y) $$ for all $x, y$. Prove that if $f(x)$ is not identically zero, and if $|f(x)| \leq 1$ for all $x$, then $|g(y)| \leq 1$ for all $y$.

Exploration

The correction cannot rely on supremum arguments or derived proportionality identities because those depend on unjustified global structure of $f$ and $g$. The only robust tool available is the functional equation itself applied along arithmetic progressions.

Fixing a real number $y$ reduces the equation to a second-order linear recurrence in the single variable $n$:

$$f((n+1)y)+f((n-1)y)=2f(ny)g(y).$$

This avoids continuity, limits, and any global boundedness arguments beyond the given $|f|\le 1$.

The key test cases $n=1,2,3,4,5$ confirm that the recurrence consistently propagates values of $f(ny)$ from any two initial values $f(0)$ and $f(y)$ without additional assumptions. The boundedness constraint must therefore force restrictions on the parameter $g(y)$.

The earlier failure came from trying to compare different values of $x$ simultaneously or introducing supremum constructions. The correct approach isolates each fixed $y$ and extracts a constraint from the recurrence itself.

Problem Understanding

The functional equation

$$f(x+y)+f(x-y)=2f(x)g(y)$$

holds for all real $x,y$, with $|f(x)|\le 1$ and $f$ not identically zero. The goal is to prove that

$$|g(y)|\le 1 \quad \text{for all } y.$$

No regularity assumptions are available, so the argument must rely entirely on algebraic consequences of the equation applied to carefully chosen inputs.

Key Observations

Substituting $x=0$ gives

$$f(y)+f(-y)=2f(0)g(y),$$

so whenever $f(0)\neq 0$, the value of $g(y)$ is immediately controlled by boundedness of $f$.

More importantly, for each fixed $y$, defining

$$a_n=f(ny)$$

produces the recurrence

$$a_{n+1}+a_{n-1}=2g(y)a_n,$$

valid for all integers $n$. This converts the problem into the study of a real second-order linear recurrence whose coefficients depend on $g(y)$.

Testing small indices confirms consistency:

for $n=1$, the recurrence is exactly the original equation with $(x,y)=(y,y)$; for $n=2,3,4,5$, it propagates values uniquely from $a_0,a_1$ without contradiction, so any growth behavior must come from the parameter $g(y)$ itself rather than from missing constraints.

Solution

Fix $y\in\mathbb{R}$ and define $a_n=f(ny)$ for all integers $n$. Substituting $x=ny$ into the functional equation gives

$$f((n+1)y)+f((n-1)y)=2f(ny)g(y),$$

hence

$$a_{n+1}+a_{n-1}=2g(y)a_n \quad \text{for all integers } n.$$

If $a_n$ is identically zero, then $f(ny)=0$ for all integers $n$. In this case, substituting $x=ny$ into the original equation yields no further restriction on $g(y)$ directly, so a different point must be used. Since $f$ is not identically zero, there exists some real $x_0$ with $f(x_0)\neq 0$, and applying the same construction to $y=x_0$ ensures a nontrivial recurrence sequence. Therefore, at least one such sequence is nonzero.

Assume now that for a given $y$ there exists some integer $n$ with $a_n\neq 0$. Then the recurrence has a nontrivial solution. Consider the characteristic equation

$$r^2-2g(y)r+1=0.$$

Its roots satisfy

$$r=g(y)\pm\sqrt{g(y)^2-1}.$$

If $|g(y)|>1$, then the discriminant is positive and the roots are distinct real numbers, one of which has absolute value strictly greater than $1$. Any nontrivial real solution of the recurrence must then contain a component growing exponentially in $|n|$, which contradicts $|a_n|\le 1$ for all integers $n$ because $|f|\le 1$ on all real inputs.

Thus for every $y$ for which the associated sequence is nontrivial, it must hold that

$$|g(y)|\le 1.$$

It remains to handle the case where $f(ny)=0$ for all integers $n$. If this occurs, then in particular $f(0)=0$ and $f(y)=0$. Substituting $x=0$ into the original equation gives

$$f(y)+f(-y)=2f(0)g(y),$$

hence $f(-y)=0$ as well. Substituting $x=y$ gives

$$f(2y)=2f(y)g(y)=0,$$

and by induction on $n$, $f(ny)=0$ for all integers $n$ is consistent but does not constrain $g(y)$.

To force a constraint in this case, choose $x$ such that $f(x)\neq 0$ (possible since $f$ is not identically zero) and apply the recurrence construction at step size $y$ starting from the shifted sequence

$$b_n=f(x+ny).$$

Then

$$b_{n+1}+b_{n-1}=2g(y)b_n$$

again holds for all integers $n$, and now $b_0=f(x)\neq 0$. Therefore the associated recurrence is nontrivial, and the previous argument applies, forcing $|g(y)|\le 1$.

Since $y$ was arbitrary, this yields

$$|g(y)|\le 1 \quad \text{for all } y.$$

This completes the proof. ∎

Verification of Key Steps

The construction of the recurrence $a_{n+1}+a_{n-1}=2g(y)a_n$ follows directly from substituting $x=ny$ into the functional equation and matches the original identity for $n=1$ through $5$, confirming consistency with small-index tests.

The characteristic equation analysis is valid because a real second-order linear recurrence with constant coefficients has general solutions governed by its roots, and when $|g(y)|>1$ at least one root has magnitude exceeding $1$, which forces unbounded growth unless the solution is identically zero.

The potential degenerate case where $f(ny)=0$ for all integers $n$ is resolved by shifting the initial condition to a point $x$ with $f(x)\neq 0$, ensuring a nontrivial recurrence sequence exists for every fixed step size $y$.

No supremum, limit process, or unjustified global proportionality is used, and each case is reduced to an explicit recurrence contradiction.

Alternative Approaches

A different approach avoids characteristic polynomials entirely by using iterated substitutions of the functional equation to generate Chebyshev-type relations

$$f((n+1)y)=2g(y)f(ny)-f((n-1)y),$$

then proving by induction that if $|g(y)|>1$, the values $f(ny)$ must eventually exceed $1$ in absolute value for any nonzero initial seed. This again contradicts $|f|\le 1$, forcing $|g(y)|\le 1$ uniformly.