functional equations with restricted domain

Question

How can I find all discontious solutions of functional equation $f(xy)=f(x)f(y)$ on $[0,1]$. Similar question is to find all solutions of the equation $f(x+y)=f(x)+f(y)$ on $[0,\infty)$. Can we still use Hamel basis?

Answer 1

To start with your second question, f is a linear operator, which in this case means a linear function with f(0) = 0. The so entire solution set is f(x) = ax where a is a constant.

Pete L. Clark provides a link below showing that there are other solutions. This link seems clearer: http://sunejakobsen.wordpress.com/2010/12/20/cauchy%E2%80%99s-functional-equation-i/ , especially in being explicit that these strange solutions are discontinuous.

The obvious continuous solutions of f(xy) = f(x)f(y) are all of the form f(x) = $x^r$ where r is a constant.

Answer 2

In certain sense, the first problem $$f(xy) = f(x)f(y)\quad\text{ for all } (x,y) \in [0,1]^2\tag{*1}$$ has one and only one more solution than corresponding problem over $[0,\infty)$: $$g(xy) = g(x)g(y)\quad\text{ for all } (x,y) \in [0,\infty)^2\tag{*2}$$

1. If $f(0) \ne 0$, then $f(0) = f(0\cdot x) = f(0)f(x) \implies f(x) = 1$ for all $x \in [0,1]$.
   It is clear $g(x) \equiv 1$ is also a solution of $(*2)$

2. If $f(1) \ne 1$, then $f(x) = f(1\cdot x) = f(1)f(x) \implies f(x) = 0$ for all $x \in [0,1]$.
   it is clear $g(x) \equiv 0$ is also a solution for $(*2)$.

3. If $f(0) = 0, f(1) = 1$ and $f(x_0) = 0$ for some $x_0 \in (0,1)$, then

   • for any $x \in (0,x_0)$, we have $f(x) = f(x_0\cdot\frac{x}{x_0}) = f(x_0)f(\frac{x}{x_0}) = 0$.
   • for any $x \in ( x_0, 1) $, we can pick a $n \in \mathbb{Z}_{+}$ such that $x^n < x_0$, we have $f(x)^n = f(x^n) = 0 \implies f(x) = 0$.

   Combine these, one find $(*1)$ has a discontinuous solution $$f(x) = \begin{cases}0, & x \in [0,1)\\1,& x = 1\end{cases}$$ which isn't a restriction for any solution of $(*2)$.

4. If a solution of $(*1)$ doesn't fall into above 3 cases, we have $f(0) = 0, f(1) = 1$ and $f(x) \ne 0,\;\forall x \in (0,1]$. We can then extend it to a function $g(x)$ over $[0,\infty)$ by: $$g(x) = \begin{cases}f(x), & x \in [0,1]\\ \frac{1}{f(\frac{1}{x})}, & x \in [1,\infty)\end{cases}$$ One can verify this $g(x)$ is well defined and is a solution of $(*2)$:

   • $x \le 1, y \le 1 \implies\\\quad\quad g(xy) = f(xy) = f(x)f(y) = g(x)g(y)$.
   • $x \le 1, y \ge 1, xy \le 1 \implies\\\quad\quad g(xy) = f(xy) = f(xy)f(\frac{1}{y})g(y) = f(x)g(y) = g(x)g(y)$.
   • $x \le 1, y \ge 1, xy \ge 1 \implies\\\quad\quad g(xy) = \frac{1}{f(\frac{1}{xy})} = f(x)\frac{1}{f(x)f(\frac{1}{xy})} = f(x)\frac{1}{f(\frac{1}{y})} = g(x)g(y) $
   • $x \ge 1, y \ge 1 \implies\\\quad\quad g(xy) = \frac{1}{f(\frac{1}{xy})} = \frac{1}{f(\frac{1}{x})f(\frac{1}{y})} = g(x)g(y)$.

What this means is aside from the $3^{th}$ case, every solution of $(*1)$ is a restriction of a solution of $(*2)$. So restricting our first problem to a restricted domain doesn't generate anything really interesting.