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$$\prime$$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.

2005-2018 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJax