How find this function equation $(f(x))^2-(f(y))^2=f(x+y)\cdot f(x-y)$

Question

Find all the continuous bounded functions $f: \mathbb R \to \mathbb R$ such that satisfying the function equation $$(f(x))^2-(f(y))^2=f(x+y)\cdot f(x-y)$$

By the way :I have see this problem( is different my problem)

if $f(x)$ is polynomial，and such

$$(f(x))^2-(f(y))^2=f(x+y)\cdot f(x-y)$$

we can prove $f$ must $f(x)=ax$

But for my problem,follow is my work firt,we set $x=y=0$ then $f(0)=0$ and let $$x=\dfrac{u+x}{2},y=\dfrac{u-x}{2}$$ we have $$f(u)f(x)=f^2(\dfrac{u+x}{2})-f^2(\dfrac{u-x}{2})$$ and let $$x=\dfrac{u-x}{2},y=\dfrac{u+x}{2}$$ then $$f(u)f(-x)=f^2(\dfrac{u-x}{2})-f^2(\dfrac{u+x}{2})=-f(u)f(x)$$ so $$f(x)=-f(-x)$$ so $f(x)$ is odd.

then I can't.Thank you

Edit: This is IMO1990 Longlisted problem POL4.

Answer 1

If $x=y=t/2$, then $f(0)f(t)=0$, so either $f(0)=0$ or $f=0$. From now on we exclude the latter case and assume that $f(0)=0$

I'll assume that $f\in C^2(\mathbb{R})$, then $$ \partial_x\partial_y(f(x+y)f(x-y))=\partial_x\partial_y(f(x)^2-f(y)^2) $$ After routine calculations we get $$ \frac{f''(x+y)}{f(x+y)}=\frac{f''(x-y)}{f(x-y)} $$ This is posible iff both fractions are constant, hence we get a differential equation $$ f''(x)-c f(x)=0 $$ for some $c\in\mathbb{R}$. Its solution is straightforward $$ f(x)= \begin{cases} Ax+B&\quad c=0\\ Ce^{x\sqrt{c}}+De^{-x\sqrt{c}}&\quad c>0\\ E\cos{x\sqrt{-c}}+F\sin{x\sqrt{-c}}&\quad c<0\\ \end{cases} $$ Recall that $f(0)=0$, then $B=0$, $C=-D$ and $E=0$. So the only possible twice differentiable solutions are $$ kx,\qquad k\sinh(xm),\qquad k\sin(xm) $$

Answer 2

We consider the functional equation $$\tag1f(x)^2-f(y)^2=f(x+y)f(x-y)$$ for all $x,y\in\mathbb R$. Using straightforward calculation (esp. the addition theorems) one verifies that the following functions are among the solutions for $(1)$:

• $f(x)=\kappa x$
• $f(x)=\kappa\sin\omega x$
• $f(x)=\kappa\sinh \omega x$

We shall call these solutions Norbert-solutions as Norbert showed in his answer that they make up all $C^2$ solutions of $(1)$. However, the problem statement was not about twice differentiable solutions, but rather about bounded continuous solutions. We shall see that all bounded continuous solutions are among the Norbert-solutions.

Let $f$ be a bounded continuous solution of $(1)$ that is not identically zero. As has been pointed out $f(0)^2-f(0)^2=f(0)f(0)$ implies that $f(0)=0$. Let $P\subseteq \mathbb R$ denote the set of periods of $|f|$. Because $f(0)=0$, we have $P\subseteq f^{-1}(0)$. On the other hand, if $a\in f^{-1}(0)$ then $f(a+x)^2-f(x)^2=f(a+2x)f(a)=0$ implies $|f(a+x)|=|f(x)|$ for all $x$, that is $a\in P$. In other words: $P=f^{-1}(0)$. As $P$ is a proper (because $f$ is not identically zero) closed (because $f$ is continuous) subgroup (always) of $\mathbb R$, we have $f^{-1}(0)= p\mathbb Z$ for some $p\ge 0$.

If $a\in P$ then also $2a\in P$, hence $f(a+x)^2-f(a-x)^2=f(2a)f(2x)=0$ and $f(a+x)f(a-x)=f(a)^2-f(x)^2\le 0 $, i.e. $$ \tag 2f(a-x)=-f(a+x)\qquad\text{if }f(a)=0.$$

If $p>0$, let $I=(0,\frac12p)$; otherwise let $I=(0,\infty)$. For $h\in I$ consider the sequence $a_k:=f(kh)$, $k\in \mathbb N$. By choice of $I$ and the IVT, $a_1=f(h)$ and $a_2=f(2h)$ have the same sign, i.e. $\frac{a_2}{2a_1}>0$. The sequence $(a_k)_{k=1}^\infty$ is uniquely determined by $a_1,a_2$ because from $(1)$ and $(2)$ we obtain the recursion $$\tag3 a_{k+1}=\begin{cases}\frac{a_k^2-a_1^2}{a_{k-1}}&\text{if }a_{k-1}\ne0,\\ -a_{k-3}&\text{if }a_{k-1}=0\end{cases}$$ for $k\ge 2$. (Note that $a_{k-1}= 0$ implies $k\ge 4$).

There exists a Norbert-solution $g\in C^2(\mathbb R)$ with $g(h)=a_1$, and $g(2h)=a_2$ (this uses the "angle duplication formulas" for $\sin$ and $\sinh$; we do not require differential calculus as one might simply verify immediately that these functions are solutions of $(1)$):

• (Parabolic case) If $\frac{a_2}{2a_1}=1$, let $g(x)=\kappa x$ with $\kappa=\frac{a_1}h $.
• (Elliptic case) If $0<\frac{a_2}{2a_1}<1$ let $g(x)=\kappa \sin\omega x$ with $\omega=\frac1h \arccos\frac{a_2}{2a_1}$ and $\kappa=\frac{2a_1\cdot |a_1|}{\sqrt{4a_1^2-a_2^2}}$
• (Hyperbolic case) If $\frac{a_2}{2a_1}>1$ let $g(x)=\kappa \sinh\omega x$ with $\omega=\frac1h \operatorname{arcosh}\frac{a_2}{2a_1}$ and $\kappa=\frac{2a_1\cdot |a_1|}{\sqrt{a_2^2-4a_1^2}}$

By induction, $g(kh)=a_k$ for all $k\in\mathbb N$. In the parabolic and hyperbolic case, the sequence $a_k$ is therefore unbounded, contrary to the assumption that $f$ is bounded. Thus $a_k=\kappa \sin k\omega h$ for all $k\in\mathbb N$. The angle $\omega h=\arccos\frac{a_2}{2a_1}$ is in $(0\frac\pi 2)$, hence the sine certainly switches signs. Therefore $f$ has nontrivial zeroes, i.e. $p>0$.

A priori, $\kappa$ and $\omega$ depend on $h$. If picking $h'=\frac 12 h$ leads to $\omega', \kappa'$, then $\omega'$ must be "compatible" with $\omega$, that is $\cos h\omega'=\cos h\omega$ and hence $2h'\omega'=h\omega'=\pm h\omega+2m\pi$. By the properties of $\arccos$, we have $0< h'\omega<\frac \pi2$ and $0<h\omega<\frac\pi 2$ so that we can conclude $\omega'=\omega$. Then automatically also $\kappa'=\kappa$. This means that the Norbert-solution $g(x)=\kappa\sin\omega x$ that we picked for $h$ coincides with $f$ not only at all $x\in h\mathbb Z$, but also for $x\in \frac12h\mathbb Z$ and by induction for all $x\in\frac1{2^n}h\mathbb Z$ and ultimately on the dense set $\bigcup_{n\in\mathbb N}\frac1{2^n}h\mathbb Z$. By continuity, $f=g$.

Remark: We dropped the parabolic and hyperbolic cases only because the problem statement asked for bounded solutions. A quick thought about these cases shows that we again have uniqueness of $\omega$ and $\kappa$ in these cases. Therefore, all continuous solutions of $(1)$ are Norbert-solutions.