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Take the functional equation $$f(x+y)=f(x)f(y).$$ This is a variation on the Cauchy equation, and we know that it has a unique continuous solution over the real numbers $$f(x)=e^{\alpha x}.$$ We then look at the pair of equations $$u(x+y)=u(x)u(y)-v(x)v(y),$$ $$v(x+y)=v(x)u(y)+u(x)v(y).$$ Moving over to complex numbers and setting $f(z)=u(z)+iv(z)$, this becomes the exponential equation above. Fixing that $u,v$ are real for real arguments fixes the trigonometric functions,

$$u(x)=e^{\alpha x}\cos\beta x,\quad v(x)=e^{\alpha x}\sin\beta x$$

as defined via complex exponentials. Further specifying a bounded image gives us what we need. So the trig functions must be the only continuous bounded solutions to the pair of equations above. Specifying that it's just bounded is probably enough, actually, given the Cauchy-ness involved.

Can this uniqueness be established by working solely over the reals? Preferably just by elementary methods. It doesn't need to be established that $\cos x=\Re e^{ix}$, for example.

EDIT:

A possible direction assuming continuity could be the following. As pointed out by an answer below, we can focus only on solutions for which $u^2+v^2=1$. With that in mind, what would be interesting to show is that given two nontrivial continuous solutions $(u_1,v_1)$ and $(u_2,v_2)$, we must have $$u_2(x)=u_1(\alpha x),$$ $$v_2(x)=v_1(\alpha x),$$ where $$\alpha=\lim_{x\rightarrow0}\frac{v_2(x)}{v_1(x)},$$ provided of course that we can show this limit exists.

Gabriel Golfetti
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1 Answers1

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What follows is just and idea.

Assume $u$ and $v$ are continuous. Let $g(x)=u^2(x)+v^2(x).$ Then $g(x+y)=g(x)g(y).$ Hence a $g(x)=e^{2\alpha x}$ for some $\alpha\in \mathbb{R}.$ For fixed $x$ there is a unique $t_x\in [0,2\pi)$ such that $$\cos(t_x)={u(x)\over \sqrt{g(x)}},\quad \sin(t_x)={v(x)\over \sqrt{g(x)}}$$ The equations involving $u(x+y)$ and $v(x+y)$ imply $$\cos(t_{x+y})=\cos(t_x+t_y),\quad \sin(t_{x+y})=\sin(t_x+t_y)$$ Hence $$t_{x+y}=t_x+t_y,\quad t_x+t_y<2\pi\\ t_{x+y}=t_x+t_y-2\pi,\quad t_x+t_y\ge 2\pi $$ Assuming continuity of $x\mapsto t_x,$ we get $t_x=\beta x.$ Hence $$u(x)=e^{\alpha x}\cos(\beta x),\qquad v(x)=e^{\alpha x}\sin(\beta x)$$

Ryszard Szwarc
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  • Some Observations : (1) When we make $f(x)$ the sum of 2 squares , we are eliminating negative values. (2) What is the justification for $t_x$ ? (3) How can we claim that it will satisfy both conditions ? – Prem Mar 13 '23 at 14:11
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    I agree with the ideas here, however my goal is to try and define the trig functions as the solution to this system of equations, so using such things as the existence of an angle with a particular $\cos$ is out of the question. – Gabriel Golfetti Mar 13 '23 at 15:32
  • @Prem $(1)$ I have modified the notation: $g=u^2+v^2.$ Clearly $g$ is positive, so we are not eliminating anything. $(2)$ If two real numbers $a$ and $b$ satisfy $a^2+b^2$ there is a unique angle $0\le \varphi<2\pi$ such that $a=\cos \varphi$ and $b=\sin\varphi.$ $(3)$ straightforward verification basing on the property $g(x+y)=g(x)g(y).$ – Ryszard Szwarc Mar 13 '23 at 15:47
  • @Prem I forgot to mention $a^2+b^2=1.$ – Ryszard Szwarc Mar 13 '23 at 18:19
  • Oh ok , It does makes sense now , @RyszardSzwarc [[ +1 ]] – Prem Mar 13 '23 at 18:51