Infinitely differentiable vs. continuously differentiable vs. analytic?

  • Thread starter fleazo
  • Start date
  • #1
81
0
Hello. I am confused about a point in complex analysis. In my book Complex Analysis by Gamelin, the definition for an analytic function is given as :


a function f(z) is analytic on the open set U if f(z) is (complex) differentiable at each point of U and the complex derivative f'(z) is continuous on U
p.45


Now, that is why I am shocked and confused to hear that there are functions that are infinitely differentiable but aren't analytic! It confuses me because we know that if a function is differentiable at a point p that means its continuous at that point. So if a function is inifinitely differentiable that means its continuous everywhere and its derivatives are continuous everywhere, right? Which is the definition of being analytic right? I have the same confusion for continuously differentiable functions which to my understanding are functions that are differentiable everywhere and their derivatives are continuous. Isn't that again all that is needed to show a function is analytic?


Thanks I am so confused!
 

Answers and Replies

  • #2
299
20
There's a difference between real analysis and complex analysis. In both real and complex analysis, a function is called analytic if it is infinitely differentiable and equal to its Taylor series in a neighborhood of every point (formally, [itex]\forall x_0 \exists \delta > 0 \forall x \vert x-x_0 \vert < \delta \Rightarrow f(x) = \sum_{j=0}^{\infty} \frac{f^{(j)}(x_0)}{j!}(x-x_0)^j[/itex]). Now, in the case of functions of a real variable, this is a stronger criterion than being infinitely differentiable. However, in the case of complex functions of a complex variable, this condition is satisfied by any function which is complex differentiable (in fact, despite your book's caution, the derivative does not need to be assumed continuous, it can be proved to be). This means that every complex-differentiable function is analytic, and so you will often have complex analysis books ignore the distinction and simply define an analytic function as one that is merely differentiable (or in your book's case, continuously differentiable), since the two concepts are equivalent in the complex variables case anyway.
 
  • #3
81
0
Thank you very much for your response. A couple of questions:


Yeah, i think in the beginning of the book they were careful to say a function that is complex diff. and continuous derivative means analytic, but later they show that if a function is analytic it is infinitely differentiable. Is the converse true? It seems to me that in the complex case, analytic <=> infinitely differentiable.


But I am confused because there is a theorem which says:

A continuously differentiable function f(z) on D is analytic <=> the differential f(z)dz is closed.


My confusion is that, i thought a continuously differentiable function would always be analytic? I'm so confused :(
 
  • #4
360
0
Hello. I am confused about a point in complex analysis. In my book Complex Analysis by Gamelin, the definition for an analytic function is given as :

a function f(z) is analytic on the open set U if f(z) is (complex) differentiable at each point of U and the complex derivative f'(z) is continuous on U
p.45
I really dislike this definition. This should be the definition for a holomorphic function. It's a http://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions" [Broken] discusses this as well:
The term analytic function is often used interchangeably with “holomorphic function”, although the word “analytic” is also used in a broader sense to describe any function (real, complex, or of more general type) that is equal to its Taylor series in a neighborhood of each point in its domain. The fact that the class of complex analytic functions coincides with the class of holomorphic functions is a major theorem in complex analysis.
 
Last edited by a moderator:
  • #5
81
0
thanks for the reply. yeah i had noticed that there was this term holomorphic which seems to be of really big importance in complex analysis. holomorphic is not actually mentioned in gamelin's book, i'm not sure why.


by the way i cleared up the confusion i had earlier about infinitely differentiable v. analytic, etc. So somebody please correct me if I'm wrong but if we have a compelx function which we can represent as:

f(z) = u + iv

and if the derivatives of u and v exist, then f is differentiable. However for complex differentiable we use the formula like we do in real analysis

lim
z --> z_{0} f(z-z_{0})-f(z)/z-z_{0} that limit should exist to be considered complex differentiable.


However, if f(z) satisfies the cauchy reimann equations, we know it's analytic, and analytic implies infinitely differentiable (Regular, not complex)


is this kind of the gist of it?

also if anyone has another complex analysis text which they are fond of please do recommend. i am not 100% happy with gamelin's book, and my prof has mentioned there are a few theorems which he does not like the way they are phrased. i really want to make sure i have a good fundamental grasp on complex analysis
 
  • #6
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
there are many good complex analysis books, but my favorite is by henri cartan.
 
  • #7
299
20
That's closer, but I think you're still getting some of your implications mixed up. We don't prove that having differentiable components that satisfy the Cauchy-Riemann equations (hereafter the Cauchy-Riemann criterion) implies the function is analytic directly. We prove that the Cauchy-Riemann criterion implies the function is holomorphic (i.e. complex differentiable) first, and then use that to show the function is analytic. I've never seen a proof that the Cauchy-Riemann criterion implies analyticity that didn't go through showing the function is holomorphic first.
 
  • #8
81
0
actually this is why im starting to think my text book isnt all that good. there is no mention of holomorphism in the book, and any time we want to prove a function is analytic we just show it satisfies the cauchy reimann equations. when i looked up holomorphic functions on my own it seems like they were pretty much the same thing as analytic functions in the complex case. just out of curiousity what is the need of first showing its holomorphic then showing analytic? doesnt one imply the other? or are you saying that the cauchy reimann equations only satisfy the holomorphism but arent directly related to analycity?
 
  • #9
299
20
Right, in the complex case holomorphic functions are the same as analytic. The Cauchy-Riemann equations allow us to conclude that the limit [itex]\frac{f(z)-f(z_0)}{z-z_0}[/itex] exists, but it's not immediately obvious that this should show that f is equal to its Talyor series around z_0, or indeed even that f has a Taylor series around z_0 (after all, that implication doesn't hold when f is real). That holomorphic functions are in fact analytic is a consequence of the Cauchy integral theorem. Perhaps looking at the http://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions" [Broken] will give you a better idea of what's involved.
 
Last edited by a moderator:
  • #10
lavinia
Science Advisor
Gold Member
3,238
625
Hello. I am confused about a point in complex analysis. In my book Complex Analysis by Gamelin, the definition for an analytic function is given as :


a function f(z) is analytic on the open set U if f(z) is (complex) differentiable at each point of U and the complex derivative f'(z) is continuous on U
p.45


Now, that is why I am shocked and confused to hear that there are functions that are infinitely differentiable but aren't analytic! It confuses me because we know that if a function is differentiable at a point p that means its continuous at that point. So if a function is inifinitely differentiable that means its continuous everywhere and its derivatives are continuous everywhere, right? Which is the definition of being analytic right? I have the same confusion for continuously differentiable functions which to my understanding are functions that are differentiable everywhere and their derivatives are continuous. Isn't that again all that is needed to show a function is analytic?


Thanks I am so confused!
Differentiable just means all of the partial derivatives exist. Analytic means that the Jacobian is multiplication by a complex number.
 
  • #11
299
20
Differentiable just means all of the partial derivatives exist. Analytic means that the Jacobian is multiplication by a complex number.
This is wrong -- even in real analysis existence of the partial derivatives is weaker than differentiability. Also, as mentioned, analytic means the function is locally equal to its Taylor series.
 
  • #12
lavinia
Science Advisor
Gold Member
3,238
625
This is wrong -- even in real analysis existence of the partial derivatives is weaker than differentiability. Also, as mentioned, analytic means the function is locally equal to its Taylor series.
right I meant to say all of the directional derivatives exist. but complex analytic is what I said.
 
  • #13
pwsnafu
Science Advisor
1,080
85
Differentiable just means all of the partial derivatives exist.
http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions"
Even if we are only talking about complex analysis, http://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem" [Broken]

Analytic means that the Jacobian is multiplication by a complex number.
Jacobian being a scaler is when you map ℝ1 → ℝ1.
For complex analysis it is always a 2 by 2 matrix.

Edit:

right I meant to say all of the directional derivatives exist.
Even that is not enough. f(0,0) = 0 and [itex]f(x,y) = \frac{y^3}{x^2+y^2}[/itex] everywhere else is a counterexample.
 
Last edited by a moderator:
  • #14
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
there are several properties:
1) the partials exist and satisfy the CR equations
2) the function has one complex derivative
3) the function has infinitely many complex derivatives
4) the function is represented by a power series everywhere locally.

these are apparently successively stronger statements. but the miracle of complex analysis is that properties 2,3,4 are all equivalent (!) if they hold on an open set. in fact i think the looman- menchoff theorem even says property 1 implies the others, but i need to check that as i am not an analyst.

well according to wikipedia i need to assume also the function is continuous in 1) to get them all equivalent.

Oh yes, the confusion arises since these properties can be proved equivalent, that some people use names for one of them to denote others of them. i.e. properly the word holomorphic applies to property 2), and the word analytic applies to property 4, but since they are all equivalent, some books use the word analytic for property 2) or even 1).

It's confusing, sorry about this.
 
Last edited:

Related Threads on Infinitely differentiable vs. continuously differentiable vs. analytic?

  • Last Post
Replies
15
Views
6K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
649
Replies
14
Views
8K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
13
Views
2K
Top