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Cauchy's functional equation is the functional equation: ${\displaystyle f(x+y)=f(x)+f(y).\ }$
$\def\Q{ℚ}\def\R{ℝ}$
We prove below that any other solutions must be highly pathological functions.
In particular, it is shown that any other solution must have the property that its graph $y = f(x)$ is dense in $\R^2,$ that is, that any disk in the plane (however small) contains a point from the graph.
From this it is easy to prove the various conditions given in the introductory paragraph.
Suppose without loss of generality that $f(q) = q$ for all $q \in \Q,$ and $f(\alpha) \neq \alpha$ for some $\alpha \in \R.$
Put
$$f(\alpha) = \alpha + \delta,\ \delta \neq 0.$$
We now show how to find a point in an arbitrary disk with center $(x, y)$ and radius $r > 0$ where $x, y, r \in \Q$ and $x \neq y.$
Put $\beta = \frac{y - x}{\delta}$ and choose a non-zero rational number $0 \neq b \in \Q$ close to $\beta$ with
$$|\beta - b| < \frac{r}{2 |\delta|} \ .$$
Then choose a rational number $a \in \Q$ close to $\alpha$ with
$$|\alpha - a| < \frac{r}{2 |b|} \ .$$
Now put:
$$X = x + b (\alpha - a)$$$$Y = f(X).$$
Then using the functional equation, we get:
\begin{alignat*}{9}Y
&= f(x + b (\alpha - a)) ~&& \\
&= f(x + b\alpha - ba) ~&& \\
&= x + b f(\alpha) - b f(a) ~&& \\
&= y - \delta \beta + b f(\alpha) - b a ~&& \text{ since } f(a) = a \text{ and } x = y - \beta \delta \\
&= y - \delta \beta + b (\alpha + \delta) - b a ~&& \text{ since } f(\alpha) = \alpha + \delta \\
&= y + b (\alpha - a) - \delta (\beta - b) ~&& \\
\end{alignat*}
Because of our choices above, the point $(X, Y)$ is inside the disk.
en.wikipedia.org/wiki/Cauchy |
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