Cauchy's functional equation

Cauchy's functional equation is one of the simplest functional equations to represent, however its solution over the real numbers is extremely complicated. The equation is

:$f(x+y)=f(x)+f(y).$

Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions $f(x)\; =\; cx$ for any arbitrary constant $c$.This family of solutions applies over the reals also. Further constraints on "f" may preclude other solutions, for example:
* if "f" is continuous (proven by Cauchy in 1821). This condition was weakened in 1875 by Darboux who showed that it was only necessary for the function to be continuous at one point.
* if "f" is monotonic on any interval.
* if "f" is bounded on any interval.On the other hand, if no further conditions are imposed on "f", then (assuming the axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by Georg Hamel using Hamel bases. The fifth problem on Hilbert's list is a generalisation of this equation.

Proof of solution over rationals

First put $y\; =\; 0$:

:$f(x+0)\; =\; f(x)\; +\; f(0)$:$f(0)\; =\; 0$

Then put $y\; =\; -x$:

:$f(x-x)\; =\; f(x)\; +\; f(-x)$:$f(-x)\; =\; -f(x)$

Then by repeated application of the function equation to $f(n\; x)\; =\; f(x\; +\; x\; +\; cdots\; +\; x)$ we get:

:$f(nx)\; =\; n\; f(x)$

And by putting $y\; =\; nx$:

:$f\; left(\; frac\{x\}\{n\}\; ight)\; =\; frac\{1\}\{n\}\; f(x)$

Putting this all together, we get:

:$f\; left(\; alpha\; q\; ight)\; =\; q\; f(alpha)\; qquad\; forall\; q\; in\; mathbb\{Q\},\; alpha\; in\; mathbb\{R\}$

Putting $alpha\; =\; 1$ we get the unique family of solutions over $mathbb\{Q\}$.

Properties of other solutions

We prove below that any other solutions must be highly pathological functions. In particular,we show that any other solution must have the property that its graph $y\; =\; f(x)$ is
dense in $mathbb\{R\}^2$, i.e. that any circle in the plane (howeversmall) contains a point from the graph. From this it is easy to prove the various conditionsgiven in the introductory paragraph.

Suppose without loss of generality that $f(q)\; =\; q\; forall\; q\; in\; mathbb\{Q\}$,and $f(alpha)\; eq\; alpha$ for some $alpha\; in\; mathbb\{R\}$.

Then put $f(alpha)\; =\; alpha\; +\; delta,\; delta\; eq\; 0$.

We now show how to find a point in an arbitrary circle, centre $(x,y)$,radius $r$ where $x,y,r\; in\; mathbb\{Q\},\; r\; >\; 0,\; x\; eq\; y$.

Put and choose a rational number$b\; eq\; 0$ close to with:

:

Then choose a rational number $a$ close to $alpha$ with:

:

Now put:

:$X\; =\; x\; +\; b\; (alpha\; -\; a)$:$Y\; =\; f(X)$

Then using the functional equation, we get:

:$Y\; =\; f(x\; +\; b\; (alpha\; -\; a))$:$=\; x\; +\; b\; f(alpha)\; -\; b\; f(a)$:::

Because of our choices above, the point $(X,\; Y)$ is inside the circle.

Proof of the existence of other solutions

The linearity proof given above also applies to any set$alpha\; mathbb\{Q\}$, a scaled copy of the rationals.We can use this to find all solutions to the equation.Note that this method is highly non-constructive, relyingas it does on the axiom of choice.

If we assume the axiom of choice, there is a basis for the reals over $mathbb\{Q\}$ i.e. a set $A\; sub\; mathbb\{R\}$ such thatfor every real number $z$ there is a unique finite set $X\; =\; left\{\; x\_1,dots\; x\_n\; ight\}\; sub\; A$ and sequence$left(\; lambda\_i\; ight)$ in $mathbb\{Q\}$such that:

:$z=\; sum\_\{i=1\}^n\; \{\; lambda\_i\; x\_i\; \}$

Now suppose that the functional equation holds on each copy of the rationals, $x\; mathbb\{Q\},\; x\; in\; A$ with constant of proportionality "g"("x"). In other words, suppose "f"("y") = "g"("x")"y" for every "y" which is a rational multiple of "x". Then by use of the decomposition above and repeated application of the functional equation, we can obtain the value of the function for any real number:

:$f(z)\; =\; sum\_\{i=1\}^n\; \{\; g(x\_i)\; lambda\_i\; x\_i\; \}$

"f"("z") is a solution to the functional equation for any $g:\; A\; ightarrow\; mathbb\{R\}$. It is linear if and only if "g" is constant.

External links

* Solution to the Cauchy Equation [http://www.math.rutgers.edu/~useminar/cauchy.pdf Rutgers University]