orl wrote: $ f$ and $ g$ are real-valued functions defined on the real line. For all $ x$ and $ y, f(x + y) + f(x - y) = 2f(x)g(y)$. $ f$ is not identically zero and $ |f(x)|\le1$ for all $ x$. Prove that $ |g(x)|\le1$ for all $ x$. Let $ y\in\mathbb R$ : If $ g(y)=\pm 1$, the result ($ |g(y)|\leq 1$) is achieved. If $ g(y)\neq \pm 1$ : We have $ f(x+y)=2g(y)f(x)-f(x-y)$ and so $ f(x+ny)=\frac{f(x)-rf(x+y)}{1-r^2}r^n+r\frac{f(x+y)-rf(x)}{1-r^2}r^{-n}$ with $ r$ any root of $ X^2-2g(y)X+1=0$ From this, it's obvious that $ |f(x)|\leq 1$ can be true only if the roots are complex, and so if the discriminant of the quadratic $ X^2-2g(y)X+1=0$ is negative. So $ g(y)^2< 1$ Q.E.D.

An other one .. . Let's denote $ 0 < S = sup(|f(x)|) < \infty$ the superior bound of $ |f|$, so $ \forall_{x,y} ,2|f(x)|.|g(y) |\leq 2S \Longrightarrow \forall_y, S.|g(y)|\leq S \Longrightarrow \forall_y ,|g(y)|\leq 1$ ..

Diogene wrote: An other one .. . Let's denote $ 0 < S = sup(|f(x)|) < \infty$ the superior bound of $ f$, so $ \forall_{x,y} ,2|f(x)|.|g(y) |\leq 2S \Longrightarrow \forall_y, S.|g(y)|\leq S \Longrightarrow \forall_y ,|g(y)|\leq 1$ .. Yessss, quick and nice! Congrats !

Diogene wrote: An other one .. . Let's denote $ 0 < S = sup(|f(x)|) < \infty$ the superior bound of $ |f|$, so $ \forall_{x,y} ,2|f(x)|.|g(y) |\leq 2S \Longrightarrow \forall_y, S.|g(y)|\leq S \Longrightarrow \forall_y ,|g(y)|\leq 1$ .. I don't really understand your solution. Could you clarify on that? I understand that $ |f(x)| |g(y)| \leq S$, but how does it follows that $ |g(y)| \leq 1$? We don't have that $ f(x) \geq S$. :

Zhero wrote: Diogene wrote: An other one .. . Let's denote $ 0 < S = sup(|f(x)|) < \infty$ the superior bound of $ |f|$, so $ \forall_{x,y} ,2|f(x)|.|g(y) |\leq 2S \Longrightarrow \forall_y, S.|g(y)|\leq S \Longrightarrow \forall_y ,|g(y)|\leq 1$ .. I don't really understand your solution. Could you clarify on that? I understand that $ |f(x)| |g(y)| \leq S$, but how does it follows that $ |g(y)| \leq 1$? We don't have that $ f(x) \geq S$. : Ya but if we have $ |f(x)||g(y)|\le S$ and $ |f(x)|\le S$ for all x, $ |g(y)|\le 1$ for all y, otherwise $ S$ contradicts the definition of $ \sup$. Think that $ S$ is not simply any upper bound but it is the least upper bound. That's what makes Diogene's solution valid.

Oh, I see; because if we have some $ c = |g(y)| > 1$, then $ |f(x)| \leq \frac{S}{c}$ for all $ x$, with $ \frac{S}{c} < S$, a contradiction. That's a very nice solution. And thanks for explaining.

Let f take on the value 1 at k.Put x=k in given equation.We are left with g=f(k+y)+f(k-y)/2

let $ |f(z)| $ is max then $ 2|f(z)|\ge |f(z+y)|+|f(z-y)|\ge |f(z+y)+f(z-y)|=2|f(z)||g(y)| $ since $ f(z)\ne 0 $ then $ |g(x)|\le 1 $ for all $ x $

$\color{red}{\textit{\textbf{Proof:}}}$ As $|f(x)|\le 1$, $|f|$ has a least upper bound $M$. Suppose that $|g(x)|>1$, then \[2|f(x)||g(x)|=|f(x+y)+f(x-y)|\le |f(x+y)|+|f(x-y)| \le 2M\]which implies \[|f(x)|\le \frac{M}{|g(x)|}<M\]which contradicts the fact that $M$ is the least upper bound of $|f(x)|$. Thus, $|g(x)| \le 1$ for all $x$. $\quad \blacksquare$

Assume the contrary. Let $K$ be a least upper bound of $|f(x)|.$ We get that $0<|f(x)|\leq K.$ It follows from a basic result and $|g(y)|>1$ that $|f(x)||g(y)|\leq K \implies |f(x)|\leq \frac{K}{|g(y)|}=L<K,$ a contradiction.