# B Uniqueness of Analytic Functions

1. Jun 25, 2017

### jackferry

Hello, I am learning about smooth analytic functions and smooth nonanalytic functions, and I am wondering the following:
Is there a theorem that states that for any real analytic functions f and g and a point a, that if at a f=g and all of their derivatives are equal, that then f=g?

2. Jun 25, 2017

### Erland

Yes, since the two functions have the same Taylor series at a, they most be equal in a neighborhood of a.

3. Jun 26, 2017

### jackferry

Does this imply that they will be equal for their entire domain?

4. Jun 26, 2017

### Erland

Yes, if the domain is connected (which, I believe, is included in the definition of "domain"): If there was a point in the domain where the functions have different values, then there is a path from the first point to this point, parametrized by h: [0,1] -> D, say. Let s be the supremum of the subset of [0,1] for which f(n)(h(t))=g(n)(h(t)) for all n.. By continuity f(n)(h(s))=g(n)(h(s)) for all n, and then by the same argument as in my previous post, f=g in a neighborhood of h(s), and hence f(n)(h(t))=g(n)h(t) for some t>s, unless s=1. This is a contradiction (even if s=1, since f(h(1)) and g(h(1)) are unequal). Hence f=g in the entire domain.

5. Jun 26, 2017

### jackferry

Awesome, thank you very much! So this means that we can enumerate all real analytical functions by choosing a point, and "listing" all the possible values of the function at that point, and then doing the same with all of its derivatives. This seems to imply that the order of analytical functions is equal to the order of the real numbers.

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6. Jun 26, 2017

### Erland

I never thought of that, but yes, you are right! The set of analytic functions in a domain has cardinatilty c.

7. Jun 26, 2017

### jackferry

Would it be possible to extend that cardinality to smooth functions somehow, or are those too complex?

8. Jun 26, 2017

### Infrared

Even better, there are only $|\mathbb{R}|$ many continuous functions $\mathbb{R}\to\mathbb{R}$. This is because a continuous function $f:\mathbb{R}\to\mathbb{R}$ is determined by its values on $\mathbb{Q}$, so the cardinality of the set of such functions is no more than the cardinality of the set of functions $\mathbb{Q}\to\mathbb{R}$. As $\mathbb{Q}$ is countable, this latter set is in bijection with the set of sequences of real numbers, which has cardinality $|\mathbb{R}|$.

9. Jun 26, 2017

### WWGD

Maybe you can trivially say that all constant functions are Analytic, which gives you the cardinality of the Reals. But I guess you want the non-trivial ones.

10. Jun 26, 2017

### WWGD

It seems you can use Complex-Analytic functions here and count them, since the two are the same when you deal with Complex-Analytic functions. I mean the Real/complex parts of a Complex-Analytic function are Real-Analytic. It seems one can also use a Baire Category argument here too.

11. Jun 26, 2017

### jackferry

That's amazing! So a slightly tangential question: when we talk about problems, esp. Differential equations, I've heard something along the lines of "there is no general analytical solution". Why is that important? Are non-analytical solutions difficult to find? Is the challenge somehow due to the fact that they can't be approximated by a taylor series?

12. Jun 26, 2017

### WWGD

Analytical , "nice" closed-form solutions are often hard to come about , but you can have, e.g., distributional/weak solutions.

13. Jun 27, 2017

### WWGD

This is one way of proving Erland's result from post #2 : Consider the set of points S={ x in (a,b): f(x)=g(x)}. We can show this set is open ( in (a,b), which is in this case the same as in the Real line):

S= {x in (a,b): f(x)=g(x)} = (a,b)- ({ x in (a,b): f(x)>g(x)} $\cup$ {x in (a,b): f(x)<g(x) } . Each of these last two sets is closed: let h(x):f(x)-g(x) , the the two sets are of the form {$f^{-1}(0)$} , inverse image of a closed set under continuous map, and therefore closed, and so is their ( finite here) union . This means S is open and therefore the union of open intervals.

EDIT: As Erland points out, there is an error here, in that the sets on the right _are not_ of the form $f^{-1}(0)$, THO, the set {$x: f(x)=g(x)$ } is of the form {$f^{-1}(0)$} BUT my original proof _does not_ hold then. Back to the drawing board.

Last edited: Jun 27, 2017
14. Jun 27, 2017

### Erland

I don't understand why the two sets are of the form {$f^{-1}(0)$}.

15. Jun 27, 2017

### WWGD

Aaargh. Right, my bad , it is $f(x)-g(x)> 0 , f(x)-g(x)<0$. Let me correct. I meant for the case $f(x)=g(x)$

16. Jun 28, 2017

### Erland

I still don't understand why these two sets are closed. They are clearly open, but why closed?

17. Jun 28, 2017

### WWGD

Sorry, I meant the set {$x: f(x)=g(x)$} is closed, as it is of the form {$f^{-1}(0)$} , while the others ( meaning {$x: f(x) >g(x)$} and {$x: f(x)<g(x)$} ) are open, as preimages of sets of the form $(a, \infty)$. EDIT:Since the later are open, whenever $f(x_0)>g(x_0)$ , there is an interval where this holds, which makes sense, given EDIT both f,g, and therefore their difference is/are continuous. EDIT2: It follows that the latter two sets are each the union of countably many open intervals.

Last edited: Jun 28, 2017