Difference between holomoprhic and analytic functions

[Silviu]

Hello! I read in my complex analysis book that holomorphic and analytic "do not always mean the same thing", but in the complex plane they do. In which case they don't mean the same thing? More specifically what does holomoprhic function means outside the complex plane (such that you can define it in a space where it is different from analytic function)?

 

[fresh_42]

A function is analytic at a point if it can be developed into a power series at this point. This implies it can be extended to a holomorphic function at this point, which means complex differentiable. In return a holomorphic function is also analytic (Taylor series). So analytic and holomorphic means locally the same for complex functions. If a complex function is everywhere analytic, then it is also everywhere holomorphic and vice versa. Complex differentiability, i.e. the Cauchy-Riemann equations hold, is a very strong requirement which we don't have for the reals.

In general, analytic only means $f(x)=\sum_{n \in \mathbb{N}_0}a_n(x-x_0)^n$ in an open neighborhood of $x_0$. But this already implies differentiability (holomorphism).

The real function, however, $f(x)=\cases{\exp(−x^{−2}) & if x≠0 \\ 0 & if x=0}$ is everywhere smooth but not analytic at zero.

 

[WWGD]

Test functions, smooth functions with compact support, are not analytic; the obvious problem is at the point where the function is 0. You need the series to converge to 0 values on one side and non-zero values on the other side. I think the statement should be: differentiable and analytic are notthe same. A function may be differentiable at a single point, though not analytic, since being analytic requires differentiability in a ball aboutthe point. I think the norm function $|z|= \sqrt{ x^2+y^2}$ is such an example. Notice C-R is satisfied only at the origin. .

 

[fresh_42]

Test functions, smooth functions with compact support, are not analytic; the obvious problem is at the point where the function is 0. You need the series to converge to 0 values on one side and non-zero values on the other side. I think the statement should be: differentiable and analytic are notthe same. A function may be differentiable at a single point, though not analytic, since being analytic requires differentiability in a ball aboutthe point. I think the norm function $|z|= \sqrt{ x^2+y^2}$ is such an example. Notice C-R is satisfied only at the origin. .

What about the complex Taylor series for a smooth function? This gives a local power series. Wikipedia says (and this is in accordance to what I have learnt, but I'm too lazy now to look it up and quote a textbook): "Actually the attributes analytic, holomorph and regular are used as synonyms in [complex] function theory."

 

[WWGD]

What about the complex Taylor series for a smooth function? This gives a local power series. Wikipedia says (and this is in accordance to what I have learnt, but I'm too lazy now to look it up and quote a textbook): "Actually the attributes analytic, holomorph and regular are used as synonyms in [complex] function theory."

Do you mean that $f(z)=|z|$ is Analytic about the origin?

 

[fresh_42]

Do you mean that $f(z)=|z|$ is Analytic about the origin?

At first glance, yes. Not over the reals, but as complex function, yes. What should prevent me from a Taylor expansion?

Edit: I've found the proof that analytic and holomorph are locally (open disc) equivalent in a book from Henri Cartan (son of Élie Cartan). Analytic functions are always holomorph and locally the Taylor series gives the other direction.

 

[WWGD]

At first glance, yes. Not over the reals, but as complex function, yes. What should prevent me from a Taylor expansion?

Edit: I've found the proof that analytic and holomorph are locally (open disc) equivalent in a book from Henri Cartan (son of Élie Cartan). Analytic functions are always holomorph and locally the Taylor series gives the other direction.

But f(z) is not even differentiable outside of 0 (ad, as I understand it, local means "in an open set about" ), and it is trivially analytic , since $f^{(n)}(0)=0$. It is actually not even a function of a "pure" Complex variable, in that it cannot be expressed in terms of z alone ; $f(z)= \sqrt{zz^{-}}$ and we cannot get rid of the $z^{-}$ (conjugate) , while Analytic functions are all expressible in terms of z, without the use of $z^{-}$ .

EDIT :
From ://en.wikipedia.org/wiki/Holomorphic_function ,(Definition, 2nd Paragraph) " If f is complex differentiable at every point z0 in an open set U, we say that f is holomorphic on U. We say that f is holomorphic at the point z0 if it is holomorphic on some neighborhood of z0.[5] We say that f is holomorphic on some non-open set A if it is holomorphic in an open set containing A.", though some may use a different definition.

 

[fresh_42]

But f(z) is not even differentiable outside of 0 (ad, as I understand it, local means "in an open set about" ), and it is trivially analytic , since $f^{(n)}(0)=0$. It is actually not even a function of a "pure" Complex variable, in that it cannot be expressed in terms of z alone ; $f(z)= \sqrt{zz^{-}}$ and we cannot get rid of the $z^{-}$ (conjugate) , while Analytic functions are all expressible in terms of z, without the use of $z^{-}$ .

Ok, if it's not smooth / differentiable / regular / holomorphic, then it's not analytic. But this wouldn't be a counterexample then. However, why should it be not differentiable at zero? Conjugation is, multiplication is and the square root is and thus the composite of all is.

Edit: I see, the square root might be the problem. Not sure about this. It's definitely too late aka early here.

 

[WWGD]

Ok, if it's not smooth / differentiable / regular / holomorphic, then it's not analytic. But this wouldn't be a counterexample then. However, why should it be not differentiable at zero? Conjugation is, multiplication is and the square root is and thus the composite of all is.

Actually it _is_ differentiable at 0, just not Analytic anywhere, since Analytic requires ( per the quoted definition) that the function be differentiable in an open ball about a point, i.e., in a region, and $f(z)=|z|$ is not, so it is diferentiable ( at 0) but nowhere Analytic. EDIT: I had rewritten the OP's condition, which I thought was incorrect, since Analytic and Holomorphic are the same. I thought he may have meant that _Differentiable_ and Analytic were not the same, and I was addressing this.

EDIT 2: I believe that a Complex function with either a purely Real or Purely Complex part, meaning, respectively that the Complex(Real ) part is identically 0 , cannot be analytic. $f(z)=|z|= \sqrt{x^2+y^2}$ is purely Real.

 

[fresh_42]

Differentiable and holomorphic is the same. The only question is, whether the Taylor series exists at zero with a positive radius of convergence or not, which I'm not sure about. But as soon as it is differentiable, it has a Taylor series in an open disc and is such analytic at this point. So it is no counterexample if it isn't differentiable. If it is, the terms are locally the same. The only difference is, that a differentiable function doesn't necessarily has the same power series everywhere, as I understand it.

 

[WWGD]

Differentiable and holomorphic is the same. The only question is, whether the Taylor series exists at zero with a positive radius of convergence or not, which I'm not sure about. But as soon as it is differentiable, it has a Taylor series in an open disc and is such analytic at this point. So it is no counterexample if it isn't differentiable. If it is, the terms are locally the same. The only difference is, that a differentiable function doesn't necessarily has the same power series everywhere, as I understand it.

But it is not differentiable outside of 0; Cauchy-Riemann is satisfied only at z=(0,0). So, differentiable _in an open set_ and holomorphic are the same, but $f(z)=|z|$ is differentiable only at z=(0,0), thus, if you accept the quoted definition, since it is not differentiable in an open set, it is not holomorphic/analytic.

 

[fresh_42]

But it is not differentiable outside of 0; Cauchy-Riemann is satisfied only at z=(0,0). So, differentiable _in an open set_ and holomorphic are the same, but $f(z)=|z|$ is differentiable only at z=(0,0), thus, if you accept the quoted definition, since it is not differentiable in an open set, it is not holomorphic/analytic.

Yes, one needs differentiability in a neighborhood $U$. What I have difficulties with is "differentiable only at z=(0,0)". Differentiability is always a local property, so it is either differentiable in a subset of $U$ or not even at a single point. How should differentiability be defined at a single spot? Maybe I should calculate C-R to see what you mean.

 

[WWGD]

Yes, one needs differentiability in a neighborhood $U$. What I have difficulties with is "differentiable only at z=(0,0)". Differentiability is always a local property, so it is either differentiable in a subset of $U$ or not even at a single point. How should differentiability be defined at a single spot? Maybe I should calculate C-R to see what you mean.

There are ways of characterizing the sets of different sorts, e.g., points where a function is continuous ( a $G_{\delta}$ IIRC ) as well as other properties.

Still, re C-R:$f(z)=\sqrt{x^2+y^2} =U(x,y); V(x,y)==0$

$U_x= \frac{2x}{\sqrt{x^2+y^2}} :=0 \rightarrow x=0$. Similar calculation for $V_y$.

 

[WWGD]

But, yes, I agree that having a property at a single point kind of seems to violate one's intuition about differentiability, in general. But remember Complex differentiability is stringent, and I think local property means that it depends on local values, not that the property itself holds in an open set.

 

[Svein]

Going back about 50 years (thus citing from an unreliable memory): Given a compact, simply connected subset K of ℂ and an open, simply connected domain O of ℂ such that K⊂O, the exists a function B(z) which is C^∞, identically 1 on K and identically 0 on the complement of O. B(z) is, however, not analytic.

 

[WWGD]

Going back about 50 years (thus citing from an unreliable memory): Given a compact, simply connected subset K of ℂ and an open, simply connected domain O of ℂ such that K⊂O, the exists a function B(z) which is C^∞, identically 1 on K and identically 0 on the complement of O. B(z) is, however, not analytic.

I would guess that functions with compact support cannot be analytic, with the problem being the transition from zero to non-zero, generalizing the issue of test functions $C^{\infty}$ with compact support) on the Real line, where the issue is at the transition from zero to non-zero values, but I am not 100% on this.

 

[Svein]

I would guess that functions with compact support cannot be analytic, with the problem being the transition from zero to non-zero, generalizing the issue of test functions $C^{\infty}$ with compact support) on the Real line, where the issue is at the transition from zero to non-zero values, but I am not 100% on this.

Google "Bump functions".

 

[WWGD]

Google "Bump functions".

Yes, I know, they are $C^{\infty}$ but not analytic. From https://en.wikipedia.org/wiki/Bump_function:

"...
Properties and uses
...While bump functions are smooth, they cannot be analytic unless they vanish identically."

 

[Hawkeye18]

The difference between holomorphic and analytic functions is explained here: https://en.wikipedia.org/wiki/Holomorphic_function
In short, as it was already mentioned above, analytic means that in a neighborhood of each point in its domain the function can be represented as the sum of convergent power series. But what is power series? For example, on a plane one can write a complex power series
$$\sum_{n=0}^\infty a_n (z-z_0)^n$$ or a real power series $$\sum_{n=0}^\infty\sum_{k=0}^\infty a_{n,k} (x-x_0)^n (y-y_0)^k. $$ If one understands the the power series as an complex power series, then "holomorphic" and "analytic" are equivalent, it is one of the fundamental facts of complex analysis.

But if one understands power series as real power series, then "analytic" means what is often called "real analytic". And real analytic functions are not always holomorphic: take for example the function $f(x,y)=x^2 + y^2$.

 

[mathwonk]

as always with questions of this sort, the answer depends on your definitions of the terms involved. (I am talking only about the complex case here.) although the linked wikipedia article requires differentiability of an open set in order to qualify as holomorphic, by contrast the definition in the book cited above by Cartan does not, allowing a function to be holomorphic at a single point. As has been stated many times, holomorphicity on an open set is equivalent to analyticity on that open set. But there do exist functions which are differentiable at only a single point, which some call holomorphic there and some do not, which are not analytic there, since analyticity implies differentiability on a neighborhood.

The question of just how much differentiability is needed to deduce analyticity is of interest, and perhaps is most clearly expressed by using the word differentiable instead of holomorphic. There are various forms of a theorem that show that requiring differentiability in all but a rather "thin" subset of an open set actually gives differentiability hence holomorphicity in the entire region and hence also analyticity. In fact even requiring continuity, plus existence of partial derivatives that satisfy the Cauchy Riemann equations in the complement of a "thin" set within a region, actually implies holomorp[hicity in the whole region. These are called theorems of "Looman-Menchoff" type. and can be googled.

by the way, although differentiability at a point p makes little sense unless a function is defined on a neighborhood of the point, it does make sense for differentiability to hold only at p and not on any neighborhood. Note e.g. that if we consider the two graphs y = x^2 and y = -x^2, that any function defined on a nbhd of x=0 whose graph lies between these two graphs must be differentiable at x=0. It is easy to imagine such a graph made of polygonal segments with sharp peaks at all points with coordinates x = 1/n, and hence not differentiable on any nbhd of x=0. I agree the more intuitive notion is continuous differentiability, where the derivative exists on a nbhd of p and is continuous on that nbhd. Indeed some basic theorems such as the inverse function theorem have that hypothesis, which guarantees that the derivative at the point is a good approximation to the function on a nbhd.