Functional equations (Table of contents)## Cauchy Equation and Equations of Cauchy Type

The equation \(f(x+y)=f(x)+f(y)\) is called the Cauchy equation. If its domain is \(\mathbb{Q}\), it is well-known that the solution is given by \(f(x)=xf(1)\). That fact is easy to prove using mathematical induction. The next problem is simply the extention of the domain from \(\mathbb{Q}\) to \(\mathbb{R}\). With a relatively easy counter-example we can show that the solution to the Cauchy equation in this case doesn't have to be \(f(x)=xf(1)\). However there are many additional assumptions that forces the general solution to be of the described form. Namely if a function \(f\) satisfies any of the conditions:

- monotonicity on some interval of the real line;
- continuity;
- boundedness on some interval;
- positivity on the ray \(x\geq 0\);

then the general solution to the Cauchy equation \(f:\mathbb{R}\rightarrow S\) has to be \(f(x)=xf(1)\).

The following equations can be easily reduced to the Cauchy equation.

- All continuous functions \(f:\mathbb{R}\rightarrow(0,+\infty)\) satisfying \(f(x+y)=f(x)f(y)\) are of the form \(f(x)=a^x\). Namely the function \(g(x)=\log f(x)\) is continuous and satisfies the Cauchy equation.
- All continuous functions \(f: (0,+\infty)\rightarrow\mathbb{R}\) satisfying \(f(xy)=f(x)+f(y)\) are of the form \(f(x)=\log_a x\). Now the function \(g(x)=f(a^x)\) is continuous and satisfies the Cauchy equation.
- All continuous functions \(f: (0,+\infty)\rightarrow(0,+\infty)\) satisfying \(f(xy)=f(x)f(y)\) are \(f(x)=x^t\), where \(t=\log_a b\) and \(f(a)=b\). Indeed the function \(g(x)=\log f(a^x)\) is continuous and satisfies the Cauchy equation.