Holomorphic functions are my muse. As my muse, I love drawing them different ways. Allow me to frame this as though an artist talking about his muse.

A holomorphic function $f$ on the unit disk $\mathbb{D}$ is completely determined (and not only determined, represented) by $\{f^{(j)}(0)\}_{j=0}^\infty$.

$$f(z) = \sum_{j=0}^\infty f^{(j)}(0) \frac{z^j}{j!}$$

Similarly a holomorphic function $f$ on $\mathbb{D}$ is completely determined (and not only determined, represented on the inside of $\mathcal{C}$) by its values on any contour $\mathcal{C} \subset \mathbb{D}$.

$$f(z) = \int_{\mathcal{C}} \frac{f(\zeta)}{\zeta - z}\,d\zeta$$

Adding more constraints, and restricting my muse to certain poses, you can find better and more nuanced art:

If $f$ is entire, and $|f(z)| < C e^{\tau |z|^\rho}$ everywhere, for arbitrary constants $C,\rho,\tau \in \mathbb{R}^+$, then $f$ is completely determined (and not only determined, $nearly$ represented) by its zeroes $\{a_j\}_{j=1}^\infty$.

$$f(z) = e^{p(z)}\prod_{j=1}^\infty (1 - \frac{z}{a_j})e^{-\frac{z}{a_j} - \frac{z^2}{2a_j^2} -...-\frac{z^n}{na_j^n}}$$

where $n$ is the closest greatest integer to $\rho$ and $p$ is a polynomial of at most degree $n$.

My muse also has rare representations, that bring out specificity and still beauty. Thanks to Ramanujan's careful deliberations,

A holomorphic function $f$ on $\mathbb{C}_{\Re(z)>0}$, such that $|f(z)|< Ce^{\rho |\Re(z)| + \tau |\Im(z)|}$, for arbitrary constants $C, \rho, \tau \in \mathbb{R}^+$ with $\tau < \pi/2$, then $f(z)$ is completely determined (and not only determined, represented) by $f \big{|}_{\mathbb{N}}$.

$$f(z)\Gamma(1-z) = \int_0^\infty \vartheta(-x)x^{-z}\,dx$$

where $\vartheta(x) = \sum_{j=0}^\infty f(j+1) \frac{x^j}{j!}$, $\Gamma$ is the Gamma function, and $0 < \Re(z) < 1$. This can be extended to the expression

$$f(z)\Gamma(1-z) = \sum_{j=0}^\infty f(j+1)\frac{(-1)^j}{j!(j+1-z)} + \int_1^\infty \vartheta(-x)x^{-z}\,dx$$

which works for all $\mathbb{C}_{\Re(z) > 0}$.

What other instances do holomorphic functions (on any domain subject to whatever constraints) admit a unique representation theorem based on a sliver of information about the function. Slightly different than identity theorems, as these determine, but rather examples that also represent.

If need be this can be Community wiki.

Thank you and Happy New Year,

Richard Diagram


1 Answer 1


The distinction between "determined" and "represented" is not clear. Consider a function $f$ analytic in domain $U$ containing, say, $0$. The values of $f$ on a sequence $p_n$ of nonzero points with limit $0$ determine $f$. If you want a "representation", you can represent the coefficients $a_k$ of the Maclaurin series recursively by limit operations: $$ a_k = \lim_{n \to \infty} \left(p_n^{-k} f(p_n) - \sum_{j=0}^{k-1} a_j p_n^{j-k}\right) $$ and then of course $f(z) = \sum_{k=0}^\infty a_k z^k$ for $z$ within the radius of convergence.

  • $\begingroup$ + Thanks, though. This is exactly the type of answer I was looking for. Would +1 if I could. $\endgroup$ Dec 31, 2018 at 5:33
  • $\begingroup$ And to your comment. Quite frankly I would take this as a representation, and then you answered the question perfectly, by trying to provide a counterexample. Things which determine holomorphic functions can tend to represent them, but this isn't necessary (perhaps it is, but one would have to "prove" this). You just showed that $f$ can be represented by an accumulative sequence. There are countless identity theorems for holomorphic functions though, they are not all necessarily representative. – $\endgroup$ Dec 31, 2018 at 5:37

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