Real- and Complex- Analytic Functions.

[Bacle]

Hi Again:

I don't know if this is obvious or not: An analytic complex function
f(z)=u(x,y)+iv(x,y) , can be made into an analytic function
f: R^2 -->R^2, since each of u(x,y) and v(x,y) is itself a real-analytic
function, i.e., we can use a standard argument by component function.

How about in the opposite direction, i.e., we have a real-analytic
function f(x), analytic in an interval (a,b). When can we extend
f(x) into a complex-analytic function.?. I suspect , thinking of power series,
that we can use the radius of convergence to construct an analytic function, i.e.,
if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
function in |z-a|<r .

Is this correct.?

 

[Anthony]

I don't know if this is obvious or not: An analytic complex function
f(z)=u(x,y)+iv(x,y) , can be made into an analytic function
f: R^2 -->R^2, since each of u(x,y) and v(x,y) is itself a real-analytic
function, i.e., we can use a standard argument by component function.

This can be seen as follows: if $$f(z)=u(x,y)+\mathrm{i} v(x,y)$$ is complex analytic, then $$u$$ and $$v$$ are harmonic hence (real) analytic.

How about in the opposite direction, i.e., we have a real-analytic
function f(x), analytic in an interval (a,b). When can we extend
f(x) into a complex-analytic function.?. I suspect , thinking of power series,
that we can use the radius of convergence to construct an analytic function, i.e.,
if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
function in |z-a|<r .

Is this correct.?

Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.

 

[Petek]

Hi Again:

How about in the opposite direction, i.e., we have a real-analytic
function f(x), analytic in an interval (a,b). When can we extend
f(x) into a complex-analytic function.?. I suspect , thinking of power series,
that we can use the radius of convergence to construct an analytic function, i.e.,
if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
function in |z-a|<r .

Is this correct.?

Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.

While Anthony's statement is true, the answer to Bacle's question is "no". Consider

$$f(x) = \frac{1}{1 + x^2}$$

in the interval (-1, 3), Using Bacle's notation, a = 1, r = 2. Then f is real-analytic in the interval (-1, 3), but f is not complex-analytic in |z - 1| < 2 due to the poles at +-i. The function f would be complex-analytic in some disk that didn't include the poles, however.

 

[Anthony]

Apologies, I only read the first bit! That will teach me.

 

[Bacle]

PeteK, Anthony: Thnaks for your posts. Please check on this; I would appreciate

your comments:
Just a question: I guess we could look at 1/(1+x^2) on the real line as agreeing

with 1-x^2+x^4 -... =Sum_(x=0...n) x^2n on (-1,1). We can also see it as

the restriction of 1/(1+z^2) , to the real axis, so that, by the identity theorem

(using the fact that (-1,1) has limit points) , 1/(1+z^2) is the analytic extension

of ( 1/(1+x^2)) to the complex plane, and :

1/(1+z^2) = 1-z^2+z^4 -... on |z|=1

But then it seems like 1/(1+x^2) on (-1,1) can be extended to the disk |z|=1. And it seems like the above could be generalized to analytic functions defined on

the real line; by the comparison criterion, if f(x) were analytic in (-r,r) (we can

translate the function if necessary so that it converges in that interval) , then

we would have -1< |(a_nx^n)/ (a_(n-1)x^n-1)|<1 in (-r,r), which I believe

would imply that the analytic extension of f(x) to the complex plane would converge

in that same interval.?
real line

 

[Bacle]

Sorry, I forgot to ask another followup:

Anthony: I understand that if f(z)=u+iv , then both u,v are harmonic, but
how does it follow from u,v being harmonic, that each is real-analytic.?.

I understand that from harmonicity, the second derivatives would be killed,
but it seems many other mixed partials would not be killed by u (equiv. v)
being harmonic.

Would you please expand.?

Thanks.

 

[Dickfore]

What's a real analytic function?

 

[Petek]

PeteK, Anthony: Thnaks for your posts. Please check on this; I would appreciate

your comments:



Just a question: I guess we could look at 1/(1+x^2) on the real line as agreeing

with 1-x^2+x^4 -... =Sum_(x=0...n) x^2n on (-1,1). We can also see it as

the restriction of 1/(1+z^2) , to the real axis, so that, by the identity theorem

(using the fact that (-1,1) has limit points) , 1/(1+z^2) is the analytic extension

of ( 1/(1+x^2)) to the complex plane, and :

1/(1+z^2) = 1-z^2+z^4 -... on |z|=1

But then it seems like 1/(1+x^2) on (-1,1) can be extended to the disk |z|=1.


And it seems like the above could be generalized to analytic functions defined on

the real line; by the comparison criterion, if f(x) were analytic in (-r,r) (we can

translate the function if necessary so that it converges in that interval) , then

we would have -1< |(a_nx^n)/ (a_(n-1)x^n-1)|<1 in (-r,r), which I believe

would imply that the analytic extension of f(x) to the complex plane would converge

in that same interval.?
real line

First, let's consider Anthony's statement that

Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.

It would be more precise to say that "Any real analytic function on some open set on the real line can be extended to a complex analytic function on some open set of the complex plane." The function $f(x) = 1/(1 + x^2)$ shows that a real-valued function can be analytic on all of R, but not have a continuation that is analytic on all of C.

Next, in your above statement you probably mean the disk {z: |z| < 1} since the complex power series $1 - x^2 + x^4 - ...$ doesn't converge, for example, if z = 1.

Finally, in your last paragraph, do you mean something like "If f(x) is analytic in (r,r), then it has a continuation to the complex numbers that is analytic in the disk {z: |z| < r}"? If so, it not true as shown by the example of f(x) as given above and with any r > 1. If you meant something else, please restate.

 

[Petek]

What's a real analytic function?

See the Wikipedia article on analytic functions.

 

[Bacle]

Thanks, PeteK.

Would you please give a hint for why 1/(1+x^2) is real-analytic in
the interval (-1,3).? I can't see it. I can see how it agrees with:


Sum_i=0...oo x^(2n) =1-x^2+x^4-... in (-1,1)

But I don't see how it is real-analytic beyond (-1,1).

Would you please explain.?

P.S: you were right, the radius of convergence of f(z)=1-z^2+z^4-...

is 1 , and, yes, there are poles at +i, -i' my bad. Tho I think

f(z) can be continued beyond the disk.

Thanks.

 

[Hurkyl]

But I don't see how it is real-analytic beyond (-1,1).

If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2.

 

[Bacle]

"If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2. "


Yes, but that is precisely where I am stuck; I can see how :

1-x^2+x^4-x^6...

is defined on (-1,1), and it agrees with 1/(1+x^2) , but I cannot

think of any other series defined outside of (-1,1), which agrees

with 1/(1+x^2). Can you spare a paradigm.?

 

[Petek]

Do you know Taylor's Theorem and Taylor Series?

 

[Bacle]

Yes, I do know both, but I have not seen them for a few years now;
I don't know if it is more accurate to say that I knew them.

Since the functions we are working with are real C^oo ,
I guess the part of the theorem that you may be referring to is that,

I do know, e.g., that, in the real case, if the radius of convergence
of the Taylor series T(f(x)) of f(x) ) is r (let's standardize to (-r,r)
again, by translation) , then T(f(x)) does not converge to f(x)
either at r, or at -r , or both. Same goes for the complex case; there
must be a point of non-convergence in the boundary of the circle of
convergence, like in the case of 1/(1+z^2) , where you pointed out
that there are poles at +i and at -i . (tho we may be able to continue
f(z) analytically in C-{i,-i} . )

This is the property I was using for 1/(1+x^2) , which has radius of
convergence one, i.e., it converges in (-1,1), and diverges at x=1 and
at x=-1 .

 

[Hurkyl]

The point is that a disk centered on 0 is probably not the best choice if you're looking for a disk that contains 2.