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F'(z)=0 implies f is not 1-1 near z

  1. Nov 29, 2007 #1
    By real or complex differentiable, I mean differentiable in an open set.

    A simple example that distinguishes real analysis from complex analysis, is the following

    Fact: f'(z) = 0 implies f is not 1-1 in any neighborhood of z.

    This is not true in the real case. For [tex]f(x) = x^3[/tex], [tex]f'(0)=0[/tex], but f is 1-1 on all of R. But for the complex function [tex]f(z)=z^3[/tex], you have [tex]f'(0)=0[/tex] but f is not 1-1 in any neighborhood of 0. To see this, for any e > 0, choose x such that 0 < x < e. Then set [tex]w = x \cdot e^{i \frac{2\pi}{3}}[/tex], and you get [tex]f(w)=w^3=x^3=f(x)[/tex]. But clearly x and w are not equal, and both x,w are in B(0,e).

    Anyways, I know it's a fact, now I need to prove it (the "fact"). If anyone is up late and knows the proof, then please do post it. Otherwise I will try to figure it out and post it...
    Last edited: Nov 29, 2007
  2. jcsd
  3. Nov 29, 2007 #2
    Good problem! I've never thought about this, and got impressed once I convinced myself, by drawing some pictures, that this seems to be true. I didn't get a proof done, and I was left with some open questions (which are probably easy, but I just don't know enough complex analysis to tell), but I'll throw something at this.

    There's a theorem, although I don't if it has any particular name, that says that if a function [itex]f:G\to\mathbb{C}[/itex], where [itex]G\subset\mathbb{C}[/itex] is open, is continuous in G, and analytic in [itex]G\backslash\{z_0\}[/itex], where [itex]z_0\in G[/itex] is some point, then f is analytic in G.

    Is it possible that an analytic function would not be open? I'm not sure. I'll assume they are open now. If it is not true always, then it's a gap in the proof.

    So let us assume that [itex]f'(z_0)=0[/itex].

    If there exists a sequence [itex]z_n\to z_0[/itex] so that [itex]f'(z_n)=0[/itex] for all n, then I don't know what to do. But if such sequence does not exist, then the proof should be available.

    Assume that there existed such ball [itex]B(z_0,r)[/itex] where f would be injective. We can assume r to be so small that [itex]f'(z)\neq 0[/itex] for all [itex]z\in B(z_0,r)\backslash\{z_0\}[/itex]. We then obtain a continuous function [itex]f^{-1}:f(B(z_0,r))\to B(z_0,r)[/itex], where [itex]f(B(z_0,r))[/itex] is some open set. According to the assumption, [itex](f^{-1})'\neq \infty[/itex] exists everywhere in [itex]f(B(z_0,r))\backslash\{f(z_0)\}[/itex], but then it follows, according to the mentioned theorem, that [itex](f^{-1})'[/itex] exists also in [itex]\{f(z_0)\}[/itex], which is a contradiction.

    That's a quick proof attempt. Hopefully it is leading to the complete proof.
    Last edited: Nov 29, 2007
  4. Nov 29, 2007 #3
    This is the only part that I didn't know for sure was true. Although I knew beforehand it's true for the real case: if [tex]\lim_{x \rightarrow c} f'(x)[/tex] exists then f'(c) exists and is that limit (assuming f is continuous at c). So I have to check this, but it seems like you can use the Cauchy Riemann equations in a similar manner to the real case.

    The "open mapping theorem": if f is nonconstant analytic on open G, then f is an open map (with respect to G). This implies that if an inverse [tex]f^{-1}[/tex] exists on f(G), then it's continuous.

    A theorem that doesn't exactly have a name is, if the set [tex] \{ z : f(z) = g(z) \}[/tex] has a limit point in a connected open set, (where f, g are analytic), then f(z)=g(z) everywhere on that open set. In this case, your hypothesis is satisfied because otherwise f'(z)=0 everywhere on the neighborhood, in which case of course f is (constant) not 1-1.

    I think it's right. I think I'm going to try to check the first claim above, using the Cauchy Riemann equations.

    You mentioned "I just don't know enough complex analysis to tell".. I just started this book Conway's Complex Analysis a couple days ago, so I'm new to it too. I've known a couple of facts, and I'm not afraid to look ahead if there is a related problem. But I have been surprised at how really good the book is so far. Of course, I skimmed (at best) most of the "definitions of complex numbers", "metric space" stuff because I already have seen that. Beyond that, the book seems to have gotten very interesting very fast. If you like this problem, it's barely scratching the surface!!
  5. Nov 29, 2007 #4
    I need to CLARIFY:

    In particular, f(x) = |x| is differentiable in [tex](-1,0) \cup (0,1)[/tex] and continuous in (-1,1), but of course f'(0) does not exist. So the claim above does NOT hold in the real case. The comment I made adds one more hypothesis: if [tex]\lim_{x \rightarrow c} f'(x) = L[/tex] exists, then in that case f'(c) does exist. You apply the mean value theorem: [tex]\frac{f(c+h)-f(c)}{h} = f'(X)[/tex], where X is in (c, c+h). Then as [tex] h \rightarrow 0^+[/tex] you have [tex]\frac{f(c+h)-f(c)}{h} = f'(X) \rightarrow L[/tex], and the same on the left, so that f'(c) = L.

    So given the claim you mention above is true, this would highlight another difference between real and complex analysis. In particular, f(z) = |z| is not an analytic function. Any analytic function that I can come up with that is "continuous but not differentiable at a point" -- i.e. has a corner -- ends up being some logarithm in which it is not defined at that point. So evidently I get the impression this is an important concept.
  6. Dec 1, 2007 #5
    I found this proved in Rudin's Real and Complex Analysis.
    Thm 10.13, Cauchy's Theorem for a Triangle, states:

    If [tex]\triangle[/tex] is a closed triangle in an open set [tex]\Omega[/tex], [tex]p \in \Omega[/tex], f is continuous on [tex]\Omega[/tex], and f is analytic on [tex]\Omega \backslash \{ p \}[/tex], then [tex]\int_{\partial \triangle} f(z) dz = 0[/tex].

    The result then follows from Morera's Theorem: http://planetmath.org/encyclopedia/MorerasTheorem.html

    I also saw this same issue discussed in the following link:

    Anyways, that's where this result comes from, I think.
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