Hölder's theorem

For every $n\; \backslash in\; \backslash N\_0,$ there is no non-zero polynomial $P\; \backslash in\; \backslash Complex\; [X;Y\_0,Y\_1,\backslash ldots,Y\_n]$ such that

$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_\{\backslash leq\; 0\}:\; \backslash qquad\; P\; \backslash left(\; z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,\{\backslash Gamma^\{(n)\}\}(z)\; \backslash right)\; =\; 0,$$

where $\backslash Gamma$ is the gamma function.

For example, define $P\; \backslash in\; \backslash Complex\; [X;Y\_0,Y\_1,Y\_2]$ by

$$P\; ~\; \backslash stackrel\{\backslash text\{df\}\}\{=\}\; ~\; X^2\; Y\_2\; +\; X\; Y\_1\; +\; (X^2\; -\; \backslash nu^2)\; Y\_0.$$

Then the equation

$$P\; \backslash left\; (z;f(z),f\text{'}(z),f$$(z) \right ) = z^2 f(z) + z f'(z) + \left (z^2 - \nu^2 \right ) f(z) \equiv 0

is called an algebraic differential equation, which, in this case, has the solutions $f\; =\; J\_\{\backslash nu\}$ and $f\; =\; Y\_\{\backslash nu\}$ — the Bessel functions of the first and second kind respectively. Hence, we say that $J\_\{\backslash nu\}$ and $Y\_\{\backslash nu\}$ are differentially algebraic (also algebraically transcendental). Most of the familiar special functions of mathematical physics are differentially algebraic. All algebraic combinations of differentially algebraic functions are differentially algebraic. Furthermore, all compositions of differentially algebraic functions are differentially algebraic. Hölder's Theorem simply states that the gamma function, $\backslash Gamma$, is not differentially algebraic and is therefore transcendentally transcendental.

Let $n\; \backslash in\; \backslash N\_0,$ and assume that a non-zero polynomial $P\; \backslash in\; \backslash Complex\; [X;Y\_0,Y\_1,\backslash ldots,Y\_n]$ exists such that

$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_\{\backslash leq\; 0\}:\; \backslash qquad\; P\; \backslash left(\; z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,\{\backslash Gamma^\{(n)\}\}(z)\; \backslash right)\; =\; 0.$$

As a non-zero polynomial in $\backslash Complex\; [X]$ can never give rise to the zero function on any non-empty open domain of $\backslash Complex$ (by the fundamental theorem of algebra), we may suppose, without loss of generality, that $P$ contains a monomial term having a non-zero power of one of the indeterminates $Y\_0,Y\_1,\backslash ldots,Y\_n$.

Assume also that $P$ has the lowest possible overall degree with respect to the lexicographic ordering For example,

because the highest power of $Y\_\{0\}$ in any monomial term of the first polynomial is smaller than that of the second polynomial.

Next, observe that for all $z\; \backslash in\; \backslash Complex\; \backslash smallsetminus\; \backslash Z\; \_\{\backslash leq\; 0\}$ we have:

$$\backslash begin\{align\}$$

&P \left(z + 1; \Gamma(z + 1), \Gamma'(z + 1), \Gamma''(z + 1),\ldots, \Gamma^{(n)}(z + 1) \right) \\[1ex]

&= P \left(z + 1;z \Gamma(z), [z \Gamma(z)]',[z \Gamma(z)]'',\ldots,[z \Gamma(z)]^{(n)} \right) \\[1ex]

&= P \left(z + 1;z \Gamma(z), z \Gamma'(z) + \Gamma(z),z \Gamma''(z) + 2\Gamma'(z),\ldots, z {\Gamma^{(n)}}(z) + n {\Gamma^{(n - 1)}}(z) \right).

\end{align}

If we define a second polynomial $Q\; \backslash in\; \backslash Complex\; [X;Y\_0,Y\_1,\backslash ldots,Y\_n]$ by the transformation

$$Q\; ~\; \backslash stackrel\{\backslash text\{df\}\}\{=\}\; ~\; P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2\; Y\_1,\backslash ldots,X\; Y\_n\; +\; n\; Y\_\{n\; -\; 1\}),$$

then we obtain the following algebraic differential equation for $\backslash Gamma$:

$$\backslash forall\; z\; \backslash in\; \backslash Complex\; \backslash setminus\; \backslash Z\; \_\{\backslash leq\; 0\}:\; \backslash qquad\; Q\; \backslash left(\; z;\backslash Gamma(z),\backslash Gamma\text{'}(z),\backslash ldots,\{\backslash Gamma^\{(n)\}\}(z)\; \backslash right)\; \backslash equiv\; 0.$$

Furthermore, if $X^h\; Y\_0^\{h\_0\}\; Y\_1^\{h\_1\}\; \backslash cdots\; Y\_n^\{h\_n\}$ is the highest-degree monomial term in $P$, then the highest-degree monomial term in $Q$ is

$$X^\{h\; +\; h\_0\; +\; h\_1\; +\; \backslash cdots\; +\; h\_n\}\; Y\_0^\{h\_0\}\; Y\_1^\{h\_1\}\; \backslash cdots\; Y\_n^\{h\_n\}.$$

Consequently, the polynomial

$$Q\; -\; X^\{h\_0\; +\; h\_1\; +\; \backslash cdots\; +\; h\_n\}\; P$$

has a smaller overall degree than $P$, and as it clearly gives rise to an algebraic differential equation for $\backslash Gamma$, it must be the zero polynomial by the minimality assumption on $P$. Hence, defining $R\; \backslash in\; \backslash Complex[X]$ by

$$R\; ~\; \backslash stackrel\{\backslash text\{df\}\}\{=\}\; ~\; X^\{h\_0\; +\; h\_1\; +\; \backslash cdots\; +\; h\_n\},$$

we get

$$Q=\; P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2Y\_1,\backslash ldots,X\; Y\_n\; +\; n\; Y\_\{n-1\})=\; R(X)\; \backslash cdot\; P(X;Y\_0,Y\_1,\backslash ldots,Y\_n).$$

Now, let $X\; =\; 0$ in $Q$ to obtain

$$Q(0;Y\_0,Y\_1,\backslash ldots,Y\_n)\; =\; P(1;0,Y\_0,2\; Y\_1,\backslash ldots,n\; Y\_\{n-1\})\; =\; R(0)\; \backslash cdot\; P(0;Y\_0,Y\_1,\backslash ldots,Y\_n)\; =\; 0\_\{\backslash Complex\; [Y\_0,Y\_1,\backslash ldots,Y\_n]\}.$$

A change of variables then yields

$$P(1;0,Y\_1,Y\_2,\backslash ldots,Y\_n)\; =\; 0\_\{\backslash Complex\; [Y\_0,Y\_1,\backslash ldots,Y\_n]\},$$

and an application of mathematical induction (along with a change of variables at each induction step) to the earlier expression

$$P(X\; +\; 1;X\; Y\_0,X\; Y\_1\; +\; Y\_0,X\; Y\_2\; +\; 2Y\_1,\backslash ldots,X\; Y\_n\; +\; nY\_\{n-1\})=\; R(X)\; \backslash cdot\; P(X;Y\_0,Y\_1,\backslash ldots,Y\_n)$$

reveals that

$$\backslash forall\; m\; \backslash in\; \backslash N:\; \backslash qquad\; P(m;0,Y\_1,Y\_2,\backslash ldots,Y\_n)\; =\; 0\_\{\backslash Complex\; [Y\_0,Y\_1,\backslash ldots,Y\_n]\}.$$

This is possible only if $P$ is divisible by $Y\_\{0\}$, which contradicts the minimality assumption on $P$. Therefore, no such $P$ exists, and so $\backslash Gamma$ is not differentially algebraic.Rubel, Lee A. “A Survey of Transcendentally Transcendental Functions”, The American Mathematical Monthly 96: pp. 777–788 (November 1989). {{JSTOR|2324840}}Boros, George & Moll, Victor. Irresistible Integrals, Cambridge University Press, 2004, Cambridge Books Online, 30 December 2011. {{doi|10.1017/CBO9780511617041.003}} Q.E.D.

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{{DEFAULTSORT:Holder's theorem}}

Category:Gamma and related functions

Category:Theorems in mathematical analysis