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A real additive function $f:\mathbb{R}\to \mathbb{R}$ that is continuous at one point $x_0$ is continuous everywhere and has form $f(x)=xf(1)$. Indeed $f(0)=0$, and $f(h)=f(x_0+h)-f(x_0)\to 0$ as $h\to 0$. Then $f(x+h)-f(x)=f(h)\to 0$ for every $x$. Additivity gives $f(q)=qf(1)$ for every rational number; continuity and the density of the rational numbers extend this identity to every real argument.

An arithmetic function $f$ is additive when $f(mn)=f(m)+f(n)$ for coprime integers $m,n$. Equivalently, $f(n)={\sum }_{p^k\parallel n}f(p^k)$. This concerns multiplication of integers, and differs from an additive function on a vector space.

Let an abelian group $V$ carry an action $\rho$ of a group $H$ by automorphisms, and use multiplication $(h,v)(k,w)=(hk,v+\rho (h)w)$. Comparing $(e,v)(h,w)$ and $(h,w)(e,v)$ shows that a translation is central precisely when its vector is fixed by the entire action. If $H$ is itself abelian, the whole semidirect product is abelian exactly when the action is trivial. For $H=(\mathbb{R},+)$, $V={\mathbb{R}}^2$, and determinant-one linear actions fixing the first coordinate vector, $\rho (x)=\left(\begin{array}{cc}1& r(x)\\ 0& 1\end{array}\right)$, where $r$ is an additive function. Without a regularity assumption, it need not be $r(x)=cx$.

If $f:[0,\infty )\to \mathbb{R}$ is continuous and is an additive function, then $f(t)=tf(1)$. Additivity proves this first for nonnegative rational numbers; because $f$ is continuous, the identity extends to every nonnegative real number.

A continuous real function satisfying $g(x+y)=g(x)g(y)$ is either identically zero or $g(x)=e^{cx}$ for a real constant $c$. If it is nonzero somewhere, $g(0)=1$ and $g(x)g(-x)=1$, so it never vanishes. The identity $g(x)=g(x/2{)}^2$ then makes it everywhere positive. Its real logarithm is a continuous additive function, hence linear by additive function continuous at one point is linear. Exponentiating proves the classification, and each listed function satisfies the equation.

For the additive function, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ impliesFor any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of additive function continuous at one point is linear.

Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,so $f(r)=cr$ for every rational number $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and density of the rational numbers giveConversely each real multiple of $x$ satisfies the Cauchy functional equation.

For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne 0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Nextmakes every value nonzero, andmakes every value positive. Thus $h(x)=\log g(x)$ is a continuous real additive function, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the continuous exponential functional equation:Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.

Since $\sigma$ is a group homomorphism, $\sigma (0)=I$, $\sigma (x+y)=\sigma (x)\sigma (y)$, and $\sigma (-x)=\sigma (x{)}^{-1}$. The product is closed on $\mathbb{R}\times {\mathbb{R}}^2$. Its two associative bracketings giveand coincide by the homomorphism property. The identity element is $(0,0)$, and the two-sided inverse isFor example, multiplying on the right gives $v-\sigma (x)\sigma (-x)v=0$, and multiplying on the left gives $-\sigma (-x)v+\sigma (-x)v=0$. Thus this law defines a group, specifically a semidirect product.

If it is an abelian group, comparing $(x,0)\ast (0,w)$ and $(0,w)\ast (x,0)$ gives $\sigma (x)w=w$ for every $w$. Hence $\sigma (x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore the unique action giving a commutative group isFor the final condition, compare $(0,a{e}_1)\ast (x,w)$ and $(x,w)\ast (0,a{e}_1)$. They are $(x,a{e}_1+w)$ and $(x,w+\sigma (x)a{e}_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma (x)e_1=e_1$ for every $x$. The first column of $\sigma (x)$ is then $(1,0{)}^{\mathsf{T}}$; since it lies in the special linear group, its determinant is one and its other diagonal entry is also one. ThusMatrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such additive function gives a homomorphism into $SL(2,\mathbb{R})$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the central translations in a linear semidirect product:No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.