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These are:
* \(\Gamma(x+1)=x\Gamma(x)\).
* \(\Gamma\left(\frac{x}{2}\right)\Gamma\left(\frac{x+1}{2}\right)=\frac{\sqrt{\pi }}{2^{x-1}}\Gamma(x)\) (Product formula).
* \(\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}\) (Euler’s functional equation).
Figure 2.16: Area of \(x^{2}+y^{2}\neq(x+y)^{2}\)
**Theorem 99**.: _(See [7], Theorem 6.3) The gamma function is the only continuous function that is positive for positive \(x\) and that satisfies the functional equations:_
* \(\Gamma(x+1)=x\Gamma(x)\)_._
* \(\Gamma\left(\dfrac{x}{p}\right)\Gamma\left(\dfrac{x+1}{p}\right)\Gamma \left(\dfrac{x+p-1}{p}\right)=\dfrac{(2\pi)^{(p-1)/2}}{p^{x-1/2}}\Gamma(x)\)_,_
_for all values of \(p\). Note that for \(p=2\) the equation above is the product formula._
The study of functional equations which have an easy appearance can get quite complicated; at the same time in some cases “clever” tricks are employed to solve the equations. Perhaps that is why we find questions involving functional equations among Math Olympiads problems. There are books dedicated to finding solutions of functional equations [45]. There are two directions concerning inquiries for any given functional equation, say \(\mathcal{F}(f)=0\). One direction deals with _regularity_, which is proving under some assumption on \(f\) that if \(\mathcal{F}(f)=0\), then \(f\) is regular (\(f\) is continuous, differentiable, analytic, etc.). The other direction concerns _stability_ which is to prove that if \(f\) is not a solution of \(\mathcal{F}(f)=0\), then there exists a unique function \(g\) which is “near”\(f\) and \(\mathcal{F}(g)=0\), meaning \(f\) almost solves the given functional equation.
Finding a solution to functional equations is one of the oldest topics of mathematical analysis. D’Alembert, Euler, Gauss, Cauchy, Abel, Weierstrass, Darboux, and Hilbert are among the renowned mathematicians who were concerned with methods of solutions for functional equations. Some trace the earliest works on functional equations to N. Oresme in the 14th century who was the first mathematician to use functional equations in several variables. He considered the following functional equation:
\[\dfrac{f(x_{1})-f(x_{2})}{f(x_{2})-f(x_{3})}=\dfrac{x_{1}-x_{2}}{x_{2}-x_{3}}\]
for all \(x_{1},x_{2},x_{3}\) with \(x_{1}>x_{2}>x_{3}\). If one picks the point \(x_{2}\) halfway between \(x_{1}\) and \(x_{3}\) one gets _Jensen’s functional equation_:
\[f\left(\dfrac{x_{1}+x_{3}}{2}\right)=\dfrac{f(x_{1})+f(x_{3})}{2}.\]
(For details see [2]).
### Cauchy’s Equation
It is generally accepted that Cauchy (1789-1857) brought rigor to analysis. It is in his book _Cours d’analyse de l’Ecole Royale Polytechnique_ that we first find systematic treatment of some basic functional equations. The equations\[f(x+y) =f(x)+f(y),\] \[f(xy) =f(x)+f(y),\] \[f(x+y) =f(x)f(y),\text{ and}\] \[f(xy) =f(x)f(y),\]
were studied by Cauchy and the functional equation \(f(x+y)=f(x)+f(y)\) is called _Cauchy’s functional equation_. If a function \(f:\mathbb{R}^{n}\to\mathbb{R}\) satisfies Cauchy’s functional equation for every \(x,y\in\mathbb{R}\), it is called _additive_. Using induction one can easily see that if a function \(f\) is additive, then for all \(x_{1},x_{2},\ldots,x_{n}\in\mathbb{R}\) for \(f:\mathbb{R}^{n}\to\mathbb{R}\) one also has:
\[f(x_{1}+x_{2}+\cdots+x_{n})=f(x_{1})+f(x_{2})+\cdots+f(x_{n}).\]
**Theorem 100**.: _If \(f:\mathbb{R}^{n}\to\mathbb{R}\) satisfies Cauchy’s functional equation, then_
\[f(rx)=rf(x)\]
_for all \(x\in\mathbb{R}^{n}\) and \(r\in\mathbb{Q}\)._
Proof
: Putting \(x=x_{1}=x_{2}=\cdots=x_{n}\) into \(f(x_{1}+x_{2}+\cdots+x_{n})=f(x_{1})+f(x_{2})+\cdots+f(x_{n})\), we get
\[f(nx)=nf(x).\]
Thus, if \(x=\left(\frac{m}{n}\right)t\) then \(f(nx)=f(mt)\); that is \(f\left(\frac{m}{n}t\right)=\left(\frac{m}{n}\right)f(t)\). Letting \(t=1\) and \(f(1)=c\), we obtain
\[f(x)=cx \tag{2.13} \]\]
for every positive rational \(x\).
For \(x=0\), \(f(0)=f(x+(-x))=f(x)+f(-x)=0\); thus \(f\) is an odd function. From the equation \(f\left(\frac{m}{n}\right)t=\frac{m}{n}f(t)\) we obtain
\[f(rt)=rf(t)\text{ \ for all real }t,\text{and all rational }\,r;\]
if \(t=1\), \(f(r)=cr\) for all rational \(r\). If \(f(x)\) is assumed to be continuous everywhere, then it follows by taking limits of both sides of (2.13) that
\[f(x)=cx\]
for all real \(x\).
The above theorem basically says that every additive function \(f:\mathbb{R}^{n}\to\mathbb{R}\) is a homomorphism from \(\mathbb{R}^{n}\) to \(\mathbb{R}\) both considered as linear spaces over \(\mathbb{Q}\).
Note that every linear function of the form \(f(x)=cx\) is additive, thus it is natural to ask if every additive function is a linear function of the form \(f(x)=cx\). The answer to this question is negative. However, by imposing a continuity condition on \(f\), we can show that \(f(x)=cx\).
**Theorem 101**.: _(Cauchy, 1821) If \(f:\mathbb{R}\to\mathbb{R}\) is continuous everywhere and additive, then \(f(x)=cx\)._
However, in 1875, G. Darboux observed that continuity only at a single point \(x_{0}\) is sufficient, since in this case it follows that \(f(x)\) is continuous everywhere. To see this, observe that continuity of \(f\) at \(x_{0}\) is expressed as \(\lim_{s\to x_{0}}f(s)=f(x_{0})\); then for any \(x\) we want to show \(\lim_{z\to x}f(z)=f(x)\). However,
\[\lim_{z\to x}f(z)=\lim_{z-x+x_{0}\to x_{0}}f[(z-x+x_{0})+(x-x_{0})]=\lim_{s\to x _{0}}f[s+(x-x_{0})].\]
Using the additivity of \(f\) then yields:
\[\lim_{z\to x}f(z)=\lim_{s\to x_{0}}f(s)+f(x-x_{0})=f(x_{0})+f(x-x_{0})=f(x_{0}+ x-x_{0})=f(x).\]
In the following we prove a different version of Cauchy’s theorem by assuming only boundedness of \(f\) on an interval.
**Theorem 102**.: _Let \(f\) be an additive function and suppose there is an interval \([a,b]\) on which \(f\) is bounded either from above or below. Then \(f(x)=cx\)._
Proof
: Assume \(f\) is both additive and bounded above on \([a,b]\). The proof is similar if \(f\) is bounded below on \([a,b]\). We want to show \(f(x)=cx\) for every \(x\). To accomplish this we set up a new function
\[h(x)=f(x)-cx\]
and observe its properties. First of all if we set \(c=f(1)\), then \(h(1)=0\) and \(h\) is additive. Note also that \(h(r)=0\) for every rational number for every rational number \(r\in\mathbb{Q}\). If \(n\) is a positive integer, then
\[0=h(1)=h\left(\frac{1}{n}n\right)=nh\left(\frac{1}{n}\right),\]
thus Since
\[h\left(\frac{m}{n}\right)=mh\left(\frac{1}{n}\right)=0\]
for every \(m=1,2,\dots\), \(h\) is vanishing at the positive rationals. But since \(h\) is an odd rationals. But since \(h\) is an odd function we obtain that \(h(r)=0\) for all \(r\in\mathbb{Q}\). The function \(h\) is also bounded:
\[h(x)=f(x)-cx\leq M+|c|\max(|a|,|b|)=K.\]
We can even claim that \(h(x)\leq K\) everywhere (just choose \(r\in[a-x,b-x]\) and use that \(h\) is \(r\)-periodic to conclude \(h(x)=h(x+r)\leq K\)). All we need to show now is that \(h(x)\leq 0\) everywhere. But \(h(nx)=nh(x)\leq K\) for every \(x\) and every \(n=1,2,\dots\) is true only when \(h(x)\leq 0\). This concludes the proof, since \(h(x)\) is also an odd function; \(h(-x)=-h(x)\leq 0\) for every \(x\) implies \(h(x)\geq 0\) everywhere and hence \(h(x)\equiv 0\).
Note that there were improvements of the above theorem, and one such result of Kormes in 1926 states that if we assume \(f\) is both additive and measurable, then \(f(x)=cx\).
**Example 86**.: Suppose a given function \(f:\mathbb{R}\to\mathbb{R}\) is a function which satisfies both
1. \(f(x+y)=f(x)+f(y)\),
2. \(f(xy)=f(x)f(y)\)
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