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I want to understand the relation between Holomorphism and Smoothness.

I want to elaborate the question as there are some underlying intricacies involved in the definitions:

Smooth: A function is smooth if it is infinitely differentiable at every point of its domain. ( I have mostly heard this definition when speaking of Real functions)

Holomorphism: A function is holomorphic if the function is differentiable at every point in the neighbourhood. (I have never heard this term when reading Real analysis)

Analytic: A function is analytic if its power series representation equals the value of the function at that point.

I know the subtle difference 1) Smoothness does not imply Analyticity for Real Analysis 2) Holomorphism implies Analyticity for Complex Analysis

I want to know if Holomorphism and Smoothness are one and the same thing. Is it that they are just two different notions where one is used in Complex Analysis and the other in Real Analysis.

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At a first glance, it sounds like smoothness (having all derivatives) implies holomorphy (having one derivative). However, there are two different meanings of derivative being used here. A complex function $f:\mathbb C\to \mathbb C$ is smooth if, when considered as a function $f:\mathbb R^2\to \mathbb R^2$, both component functions of the output have all higher order partial derivatives. On the other hand, a complex function $f$ is called holomorphic if it is complex differentiable, meaning its first partial derivatives exist and satisfy the Cauchy-Riemman equations.

To see these are not equivalent, note that $z\mapsto \overline{z}$ is smooth (it is a linear map $\mathbb R^2\to \mathbb R^2$), but not holomorphic.

It is true that every holomorphic function is smooth.

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  • $\begingroup$ Seems to be making sense. Can you elaborate a bit on z* example. $\endgroup$ Dec 6, 2018 at 6:31
  • $\begingroup$ Do you want me to elaborate one why it is smooth, or why it is not holomorphic? @ChetanWaghela $\endgroup$ Dec 6, 2018 at 6:33
  • $\begingroup$ If you can elaborate both it can be very helpful. $\endgroup$ Dec 6, 2018 at 6:35
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    $\begingroup$ @Chetan Write it down in $(x,y)$ coordinates and calculate the derivatives (real and complex)! This is a very enlightening exercise. $\endgroup$
    – user98602
    Dec 6, 2018 at 6:39
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    $\begingroup$ ♪ too many mikes, it's trueee ♪ $\endgroup$
    – user98602
    Dec 6, 2018 at 6:50
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We can parametrize $\mathbb{C}$ with a complex coordinate $z$ or a pair of real coordinates $x+iy$ where $z=x+iy$. This lets us look at points in $\mathbb{C}$ as pairs of real numbers $(x,y)$ without any serious issues. It turns out that "complex differentiability" of $f$ at $z$ i.e. the condition that $$ f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$$ exists is very strong. Using this we can prove that $f$ viewed as a function $f(x,y)=u(x,y)+iv(x,y)$ from $\mathbb{R}^2\to \mathbb{R}$ satisfies the Cauchy-Riemann equations: $$ \begin{cases} \partial_xu=\partial_yv\\ \partial_y u =-\partial_x v. \end{cases}$$ Developing the theory of functions of a complex variable, we can prove that any complex differentiable (holomorphic) function $f$ on an open domain $\Omega\subseteq \mathbb{C}$ is analytic. That is, it can be written as a convergent power series locally around any $z_0\in \Omega$ as $$ f(z)=\sum_{k=0}^\infty a_k (z-z_0^k).$$ Such functions are called analytic. On the other hand, there exist smooth non-analytic functions in the real case. For instance take the bump function $g:\mathbb{R}^2\to \mathbb{R}$ given by $$ g(x,y)= \begin{cases} \exp\left(\frac{-1}{1-(x^2+y^2)}\right)& x^2+y^2<1\\ 0 & \text{else}. \end{cases}$$ This function is infinitely many times differentiable (as you can check), but does not have a convergent power series expansion. The fact that such bump functions exists makes the real case of manifold theory a bit more flexible, while the complex situation is a bit more rigid. The difference is that complex differentiable functions also have to satisfy a system of partial differential equations, in essence.

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  • $\begingroup$ i posted an answer. please let me know what you think. $\endgroup$
    – BCLC
    Oct 11, 2021 at 19:19
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The shortcut:


For real of 1 variable:

Analytic $\implies$ smooth $\implies$ (real-)holomorphic $\implies$ differentiable (at a point $p$).

  • here, real holomorphic at $p$ just means real differentiable in an open neighbourhood of $p$. and then smooth at $p$ implies real differentiable not just on $p$ but on an open neighbourhood of $p$. (But analytic is necessarily for neighbourhood anyway, so ).

For complex of 1 variable:

Analytic $\iff$ holomorphic $\implies$ smooth $\implies$ differentiable (at a point $p$).

  • In either case, smooth is: $f$ is smooth at $t$ if $f^{n}(t)$ exists for each $n$.

The important thing to note: analytic and smooth are pretty much the same in complex as in real. The difference here is holomorphic.

  1. analytic and holomorphic: Of course analytic implies holomorphic (same in complex as in real), but the big deal in complex analysis is that holomorphic implies analytic. (I don't think analytic in complex is that much stronger as compared to in real. Same with smooth.)

  2. smooth and holomorphic: Holomorphic in complex is now so strong that smooth doesn't imply holomorphic anymore. Now, holomorphic implies smooth.

Basically $f'(z)$'s existence in a neighbourhood of $z$ implies the existence of each $f^{n}(z)$ in a neighbourhood (the same neighbourhood right?) of $z$. But this doesn't hold for real. We can have $f'(x)$'s existence in a neighbourhood of $x$, but we're not sure even for the 2nd derivative that $f^{2}(x)$ exists anywhere in a (a? the? any?) neighbourhood of $x$.

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