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By reading the Cauchy's Functional Equations on the Wiki, it is said that

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. Such functions are sometimes called Hamel functions.

Could anyone give a more explicit explanation of these many other solutions?

Besides the trivial solution of the form $f(x)=C x$, where $C$ is a constant, and the solution above constructed by the Hamel Basis, are there any more solutions existing?

Scorpio19891119
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  • The quoted paragraph says that infinitely many solutions can be found because there are infinitely many Hamel bases. – Tunococ Sep 08 '13 at 23:40
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    The "trivial" solutions also can be constructed from a Hamel basis. There are lots of solutions because the function can be defined arbitrarily on the elements of a Hamel basis. – André Nicolas Sep 08 '13 at 23:47

2 Answers2

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All solutions are "Hamel basis" solutions. Any solution of the functional equation is $\mathbb{Q}$-linear. Let $H$ be a Hamel basis. Given a solution $f$ of the functional equation, let $g(b)=f(b)$ for every $b\in H$, and extend by $\mathbb{Q}$-linearity. Then $f(x)=g(x)$ for all $x$.

There are lots of solutions because a Hamel basis has cardinality the cardinality $c$ of the continuum. A solution can assign arbitrary values to elements of the basis, and be extended to $\mathbb{R}$ by $\mathbb{Q}$-linearity. So there are $c^c$ solutions. There are only $c$ linear solutions, so "most" solutions of the functional equation are non-linear.

André Nicolas
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Discontinuous solutions of this equation are non-measurable, which implies, in a strong sense, the inability to construct them explicitly by the standards of classical analysis.

(Funny timing of the question. This was just explained in Hamel bases without (too much) axiomatic set theory .)

Community
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zyx
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