IMO 1972 Problem 5
The functional equation is
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m24s
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 functional equation is
$$f(x+y)+f(x-y)=2f(x)g(y) \qquad (\ast)$$
for all real $x,y$, with $|f(x)|\le 1$ everywhere and $f$ not identically zero. The goal is to prove that $|g(y)|\le 1$ everywhere.
A first attempt is to solve for $g(y)$:
$$g(y)=\frac{f(x+y)+f(x-y)}{2f(x)}$$
whenever $f(x)\neq0$.
Since the numerator has absolute value at most $2$, this gives
$$|g(y)|\le \frac1{|f(x)|}.$$
This is useless unless one can choose $x$ with $|f(x)|=1$; such a point need not exist.
The condition that $f$ is not identically zero suggests fixing a point $a$ with $f(a)\neq0$. Then
$$f(a+y)+f(a-y)=2f(a)g(y).$$
If for some $y$ we had $|g(y)|>1$, then the right-hand side would have absolute value strictly larger than $2|f(a)|$. Since each term on the left has absolute value at most $1$, the triangle inequality only yields
$$2|f(a)||g(y)|\le 2,$$
hence
$$|f(a)|\le \frac1{|g(y)|}.$$
This shows that every nonzero value of $f$ must be small whenever $|g(y)|>1$.
The next idea is to exploit the functional equation repeatedly. Fix $y$ and define
$$M=\sup_{x}|f(x)|.$$
Because $f$ is not identically zero and bounded by $1$, we have
$$0<M\le1.$$
Choose points where $|f(x)|$ is arbitrarily close to $M$. Applying $(\ast)$ at such points gives
$$2|f(x)||g(y)| = |f(x+y)+f(x-y)| \le |f(x+y)|+|f(x-y)| \le 2M.$$
Hence
$$|f(x)||g(y)|\le M.$$
Letting $|f(x)|\to M$ yields
$$M|g(y)|\le M.$$
Since $M>0$, this implies
$$|g(y)|\le1.$$
This argument is short and avoids any need to determine the structure of the solutions. The delicate point is that the supremum need not be attained, so one must use an approximating sequence.
The expected subclaims are that the supremum $M$ is positive and that a sequence approaching the supremum exists by the definition of supremum.
Problem Understanding
We are given two real-valued functions defined on all real numbers and satisfying
f(x+y)+f(x-y)=2f(x)g(y) ] for every pair of real numbers $x$ and $y$. The function $f$ is bounded by $1$ in absolute value and is not identically zero. This is a Type B problem. The statement already specifies what must be proved: namely, that
|g(y)|\le1
for every real number $y$. The objects involved are arbitrary real-valued functions satisfying a functional equation. No continuity, differentiability, or algebraic form is assumed. The difficulty is that the equation links values of both functions at different arguments, and there is no obvious way to isolate $g(y)$ because $f(x)$ may vanish at many points. A naive division by $f(x)$ is not globally valid. The key observation is that the boundedness of $f$ can be transferred to $g$ by examining points where $|f|$ is arbitrarily close to its largest possible value. ## Proof Architecture The proof uses two lemmas. Lemma 1. If
M=\sup_{x\in\mathbb R}|f(x)|,
then $M>0$. Sketch: Since $f$ is not identically zero, there exists a point where $|f|>0$, and the supremum must be at least that value. Lemma 2. For every fixed real number $y$,
M|g(y)|\le M.
Sketch: Choose a sequence $(x_n)$ with $|f(x_n)|\to M$. Apply the functional equation at $x_n$, bound the left-hand side by the triangle inequality and the definition of $M$, then pass to the limit. The hardest part is handling the possibility that the supremum is not attained. The argument must use an approximating sequence rather than a maximizing point. ## Solution Define
M=\sup_{x\in\mathbb R}|f(x)|.
Since $|f(x)|\le1$ for all $x$, the number $M$ exists and satisfies $0\le M\le1$. ### Lemma 1 The number $M$ is strictly positive. #### Proof Because $f$ is not identically zero, there exists a real number $a$ such that
f(a)\neq0.
$$Hence$$
|f(a)|>0.
Since $M$ is the supremum of the set ${|f(x)|:x\in\mathbb R}$, we have
M\ge |f(a)|>0.
Thus $M>0$. ∎ This establishes that the supremum of $|f|$ is nonzero; replacing the supremum by $0$ would contradict the existence of a point where $f$ is nonzero. ### Lemma 2 For every real number $y$,
M|g(y)|\le M.
Proof Fix an arbitrary real number $y$. By the definition of supremum, for each positive integer $n$ there exists a real number $x_n$ such that
|f(x_n)|>M-\frac1n.
$$Consequently,$$
\lim_{n\to\infty}|f(x_n)|=M.
Applying the functional equation at $(x_n,y)$ gives
f(x_n+y)+f(x_n-y)
2f(x_n)g(y).
$$Taking absolute values and using the triangle inequality,$$
2|f(x_n)||g(y)|
|f(x_n+y)+f(x_n-y)|
\le
|f(x_n+y)|+|f(x_n-y)|.
Since $M$ is the supremum of $|f|$,
|f(x_n+y)|\le M,
\qquad
|f(x_n-y)|\le M.
$$Therefore$$
2|f(x_n)||g(y)|
\le
2M,
$$or equivalently,$$
|f(x_n)||g(y)|
\le
M.
Passing to the limit as $n\to\infty$ and using
|f(x_n)|\to M,
$$we obtain$$
M|g(y)|\le M.
This proves the lemma. ∎ This establishes the crucial inequality at the level of the supremum; choosing a point where the supremum is attained would be unjustified because such a point need not exist. We now complete the proof. Let $y$ be any real number. Lemma 2 gives
M|g(y)|\le M.
By Lemma 1, $M>0$, so division by $M$ is legitimate and yields
|g(y)|\le1.
Since $y$ was arbitrary, the inequality holds for every real number $y$. This completes the proof. ∎ ## Verification of Key Steps The first delicate step is the passage from the supremum to a sequence approaching it. Let
M=\sup_{x}|f(x)|.
The definition of supremum states that for every $\varepsilon>0$ there exists $x$ with
|f(x)|>M-\varepsilon.
Choosing $\varepsilon=1/n$ produces a sequence $(x_n)$ satisfying
|f(x_n)|>M-\frac1n.
Since also $|f(x_n)|\le M$, the squeeze theorem yields
|f(x_n)|\to M.
A careless argument would assume the existence of $x$ with $|f(x)|=M$, which need not hold for an arbitrary bounded function. The second delicate step is the estimate
2|f(x_n)||g(y)|
\le 2M.
$$Starting from$$
2f(x_n)g(y)
f(x_n+y)+f(x_n-y),
$$one obtains$$
2|f(x_n)||g(y)|
|f(x_n+y)+f(x_n-y)|.
$$The triangle inequality gives$$
|f(x_n+y)+f(x_n-y)|
\le
|f(x_n+y)|+|f(x_n-y)|.
Each summand is at most $M$, so the right-hand side is at most $2M$. Omitting the triangle inequality and comparing absolute values term by term would not be valid. The third delicate step is dividing by $M$. This requires proving first that $M>0$. If one skipped Lemma 1, the division could fail when $M=0$. The condition that $f$ is not identically zero excludes this possibility. ## Alternative Approaches Another route begins by assuming that for some fixed $y_0$ one has $|g(y_0)|>1$. For every $x$ the functional equation gives
2|f(x)||g(y_0)|
|f(x+y_0)+f(x-y_0)|
\le
|f(x+y_0)|+|f(x-y_0)|
\le
$$Hence$$
|f(x)|
\le
\frac1{|g(y_0)|}
\qquad\text{for all }x.
This improves the original bound $|f(x)|\le1$. Repeating the same argument with the stronger bound shows that
|f(x)|
\le
\frac1{|g(y_0)|^2},
]
then
|f(x)| \le \frac1{|g(y_0)|^3}, ] and so on. Since $|g(y_0)|>1$, letting the number of iterations tend to infinity yields $f(x)=0$ for every $x$, contradicting the hypothesis that $f$ is not identically zero. This alternative proof is elegant but requires an iterative argument. The supremum method is preferable because it reaches the conclusion in a single application of the functional equation and avoids constructing an infinite chain of improved bounds.