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| '''Cauchy's functional equation''' is the [[functional equation]]
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| :<math> f(x+y)=f(x)+f(y). \ </math>
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| Solutions to this are called [[additive function]]s.
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| Over the [[rational numbers]], it can be shown using elementary algebra that there is a single family of solutions, namely <math> f(x) = cx \ </math> for any arbitrary rational number <math>c</math>.
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| Over the [[real number]]s, this is still a family of solutions; however there can exist other solutions that are extremely complicated. Further constraints on ''f'' sometimes preclude other solutions, for example:
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| * if ''f'' is [[continuous function|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.
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| * if ''f'' is [[monotonic function|monotonic]] on any interval.
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| * if ''f'' is [[bounded function|bounded]] on any interval.
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| 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''.<ref>Kuczma (2009), p.130</ref>
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| The [[Hilbert's fifth problem|fifth problem]] on [[Hilbert's problems|Hilbert's list]] is a generalisation of this equation. Functions where there exists a [[real number]] <math>c</math> such that <math> f(cx) \ne cf(x) \ </math> are known as Cauchy-Hamel functions and are used in Dehn-Hadwiger invariants which are used in the extension of [[Hilbert's third problem]] from 3-D to higher dimensions.<ref>V.G. Boltianskii (1978) "Hilbert's third problem", Halsted Press, Washington</ref> | |
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| == Proof of solution over rationals ==
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| First put <math>y = 0</math>:
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| :Then <math> f(x) = f(x+0) = f(x) + f(0) \ </math>
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| :Hence <math> f(0) = 0 \ </math>
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| Then put <math>y = -x</math>:
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| :Then <math> 0 = f(0) = f(x-x) = f(x) + f(-x) \ </math>
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| :Hence <math> f(-x) = -f(x) \ </math>
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| Then by repeated application of the function equation to <math> f(n x) = f(x + x + \cdots + x) </math> we get:
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| :<math> f(nx) = n f(x) \ </math>
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| And by replacing <math> x </math> with <math>\frac{x}{n}</math>:
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| :<math> f(x) = f \left( n \cdot \frac{x}{n} \right) = n f \left( \frac{x}{n} \right) \ </math>
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| Therefore
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| :<math> f \left( \frac{x}{n} \right) = \frac{1}{n} f(x) \ </math>
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| By putting the last three equations together, we get for any rational number <math>\frac{m}{n}</math>:
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| :<math> f \left( \frac{m}{n}x \right) = \frac{m}{n} f(x) \ </math>
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| Putting this all together, we get:
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| :<math> f \left( \alpha q \right) = q f(\alpha) \qquad \forall q \in \mathbb{Q}, \alpha \in \mathbb{R} \ </math>
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| Putting <math>\alpha = 1</math> we get the unique family of solutions over <math>\mathbb{Q}</math>.
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| ==Properties of other solutions==
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| We prove below that any other solutions must be highly [[Pathological (mathematics)|pathological]] functions. In particular,
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| we show that any other solution must have the property that its graph <math>y = f(x)</math> is
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| [[dense set|dense]] in <math>\mathbb{R}^2</math>, i.e. that any disk in the plane (however
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| small) contains a point from the graph. From this it is easy to prove the various conditions
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| given in the introductory paragraph.
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| Suppose without loss of generality that <math>f(q) = q \ \forall q \in \mathbb{Q}</math>,
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| and <math>f(\alpha) \neq \alpha</math> for some <math>\alpha \in \mathbb{R}</math>. | |
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| Then put <math>f(\alpha) = \alpha + \delta, \delta \neq 0</math>.
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| We now show how to find a point in an arbitrary circle, centre <math>(x,y)</math>,
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| radius <math>r</math> where <math>x,y,r \in \mathbb{Q}, r > 0, x \neq y</math>.
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| Put <math>\beta = \frac{y - x}{\delta}</math> and choose a rational number
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| <math>b\neq 0</math> close to <math>\beta</math> with:
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| :<math>\left| \beta - b \right| < \frac{r}{2 \left|\delta\right|}</math>
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| Then choose a rational number <math>a</math> close to <math>\alpha</math> with:
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| :<math>\left| \alpha - a \right| < \frac{r}{2\left|b\right|} </math>
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| Now put:
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| :<math>X = x + b (\alpha - a) \ </math> | |
| :<math> Y = f(X) \ </math>
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| Then using the functional equation, we get:
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| :<math> Y = f(x + b (\alpha - a)) \ </math>
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| :<math> = x + b f(\alpha) - b f(a) \ </math> | |
| :<math> = y - \delta \beta + b f(\alpha) - b f(a) \ </math>
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| :<math> = y - \delta \beta + b (\alpha + \delta) - b a \ </math>
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| :<math> = y + b (\alpha - a) - \delta (\beta - b) \ </math>
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| Because of our choices above, the point <math>(X, Y)</math> is inside the circle.
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| ==Proof of the existence of other solutions== | |
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| The linearity proof given above also applies to any set
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| <math>\alpha \mathbb{Q}</math>, a scaled copy of the rationals.
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| We can use this to find all solutions to the equation.
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| Note that this method is highly non-constructive, relying
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| as it does on the [[axiom of choice]].
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| If we assume the axiom of choice, there is a basis for the reals over <math>\mathbb{Q}</math>
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| i.e. a set <math>A \sub \mathbb{R}</math> such that
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| for every real number <math>z</math> there is a unique finite set
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| <math>X = \left\{ x_1,\dots x_n \right\} \sub A</math> and sequence
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| <math>\left( \lambda_i \right)</math> in <math>\mathbb{Q}</math>
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| such that:
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| :<math> z= \sum_{i=1}^n { \lambda_i x_i }</math>
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| By the argument above, on each copy of the rationals, <math>x \mathbb{Q}, x \in A</math>, <math>f </math> must coincide with a linear map, say with constant of proportionality ''g''(''x''). In other words, ''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:
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| :<math> f(z) = \sum_{i=1}^n { g(x_i) \lambda_i x_i }</math>
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| ''f''(''z'') is a solution to the functional equation for any <math>g: A \rightarrow \mathbb{R}</math>, and every solution is of this form. ''f'' is linear if and only if ''g'' is constant.
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| == External links ==
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| * Solution to the Cauchy Equation [http://www.math.rutgers.edu/~useminar/cauchy.pdf Rutgers University]
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| * [http://cofault.com/2010/01/hunt-for-addictive-monster.html The Hunt for Addi(c)tive Monster]
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| == References ==
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| {{Reflist}}
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| *{{cite book | last = Kuczma | first=Marek | authorlink=Marek Kuczma| title=An introduction to the theory of functional equations and inequalities. Cauchy's equation and Jensen's inequality | publisher=Birkhäuser | location = Basel | year=2009 | isbn=9783764387495 }}
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| {{DEFAULTSORT:Cauchy's Functional Equation}}
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| [[Category:Arithmetic functions]]
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| [[Category:Functional equations]]
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Greetings. The writer's title is Phebe and she feels comfortable when people use the full title. Hiring is my occupation. To collect cash is 1 of the things I adore most. For a while I've been in South Dakota and my mothers and fathers reside close by.
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