Intereting Posts

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This question is a possibly harder version of: Find $g'(x)$ at $x=0$.

**Question.** *Let $f,g :\mathbb R\to\mathbb R$, such that
\begin{align}
f(x-y)=f(x)\, g(y)-f(y)\, g(x), \tag{1}\\
g(x-y)=g(x)\, g(y)+f(x)\, f(y), \tag{2}
\end{align}
for all $x,y \in \mathbb{R} $. If $g$ is continuous at $x=0$ and not identically zero, then there exists an $\alpha\in\mathbb R$, such
$$
f(x)=\sin \alpha x\quad\text{and}\quad g(x)=\cos \alpha x.
$$
Is there a pair of discontinuous $f$ and $g$ satisfying $(1)$ and $(2)$*?

**Update.** If $\ell :\mathbb R\to\mathbb R$ is a linear functional over $\mathbb Q$ (i.e., $\ell(qx+ry)=q\ell(x)+r\ell(y)$, for all $x,y\in\mathbb R$ and $q,r\in\mathbb Q$), then

$$

\sin\big(\ell(x)\big), \quad \cos\big(\ell(x)\big),

$$

satisfy $(1)$ and $(2)$. Discontinuous such functionals do exist, and they are obtainable

using Zorn’s Lemma (equivalently the Axiom of Choice.) This takes care of the second question.

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Since $f(0)=0$ and $g(0)=1$, we also have $f(-y)=-f(y)$ and

\begin{align}

f(x+y)=f(x)g(y)+f(y)g(x),\\

g(x+y)=g(x)g(y)-f(x)f(y).

\end{align}

Set $\psi(x)=g(x)+if(x)$. Then

\begin{align}

\psi(x+y)=g(x+y)+if(x+y)

&=g(x)g(y)-f(x)f(y)+i(f(x)g(y)+f(y)g(x))\\[1ex]

&=(g(x)+if(x))(g(y)+if(y))\\[1ex]

&=\psi(x)\psi(y)

\end{align}

Thus $\psi$ is a homomorphism of the additive group $\mathbb{R}$ into the multiplicative group $\mathbb{C}\setminus\{0\}$. Conversely, any homomorphism from $\mathbb{R}$ to the multiplicative group $\mathbb{C}\setminus\{0\}$ provides a solution to the functional equations we’re dealing with, by taking the real and imaginary parts for $g$ and $f$ respectively.

Let $\varphi\colon\mathbb{C}\to\mathbb{C}$ be a field automorphism. Consider the map

$$

\psi\colon\mathbb{R}\to\mathbb{C},\qquad

\psi(x)=\varphi(e^x).

$$

Then $\psi$ is a homomorphism of the additive group of $\mathbb{R}$ into the group $\mathbb{C}\setminus\{0\}$.

Let’s take as $\varphi$ an automorphism that doesn’t send the reals into the reals; the existence of such automorphisms was first proved as a consequence of Steinitz’s theorem by Segre (Atti dell’Accademia dei Lincei, 1947). Of course, this requires the axiom of choice. Basically, an automorphism is defined by an arbitrary permutation of a transcendency basis of $\mathbb{C}$ over $\mathbb{Q}$. It’s sufficient to send a real element (we can always assume one is present, say $e$) into a non real one (which of course must exist).

By Theorem 2 in a paper by Kestelman (Proc. London Math. Soc. (2), 1951), the image of the reals under such an automorphism is dense in the complex numbers; since

$$

\phi(\mathbb{R})=\psi(\mathbb{R})\cup\{0\}\cup(-\psi(\mathbb{R})),

$$

also $\psi(\mathbb{R})$ must be dense in $\mathbb{C}$, so it can’t be contained in the unit circle and so $\psi$ has not the form $\psi(x)=\cos(ax)+i\sin(ax)$, for any real $a$.

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