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Constant function

  1. Oct 20, 2007 #1
    Please check whether this makes sense

    1. The problem statement, all variables and given/known data

    If [itex]U\subset\mathbb{C}[/itex] open, path-connected and [itex]f:U\longrightarrow\mathbb{C}[/itex] differentiable with [itex]f'(z)=0[/itex] for all [itex]z\in{U}[/itex], then f is constant.

    H1: U is pathconnected
    H2: [itex]f:U\longrightarrow\mathbb{C}, f'(z)=0, z\in{U} [/itex]

    2. The attempt at a solution

    By H1 for every [itex]p,q\in{U},p\neq{q}[/itex] there exists [itex]\gamma:[a,b]\longrightarrow{U}[/itex] such that [itex]\gamma(a)=p,\gamma(b)=q[/itex].
    So let [itex]\gamma[/itex] be an arbitrary path in U and let [itex]z\in\gamma([a,b])\subset{U}[/itex], that is [itex]z:=\gamma(t)[/itex] for [itex]t\in[a,b][/itex].

    We shall show that f is constant on any path in U between arbitrary but fixed points p and q. Since [itex]\gamma[/itex] is arbitrary, it will then follow that f is constant on U.

    Since f is complex-differentiable and gamma is continuous, and hence also real-differentiable, we have [itex](f\circ{\gamma})'(t)={\gamma}'(t)f'(\gamma(t))[/itex]. But [itex]f'(\gamma(t))=f'(z)=0[/itex] by H2, hence [itex](f\circ{\gamma})'(t)=0[/itex], which implies that [itex](f\circ{\gamma})(t)=const[/itex]. In particular, [itex](f\circ{\gamma})(a)=(f\circ{\gamma})(b)[/itex], that is [itex]f({\gamma}(a))=f({\gamma}(b))[/itex], or [itex]f(p)=f(q)[/itex] for all [itex]p\neq{q}, p,q\in U[/itex]. But this is exactly the property of a constant function. It follows that f is locally, on a subset of U, constant and hence constant.
    Last edited: Oct 20, 2007
  2. jcsd
  3. Oct 20, 2007 #2


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    But could you prove this?

    My hint at the solution to the problem would be: path integrals.
  4. Oct 21, 2007 #3
    Yes, I think I can.

    We have [itex](f\circ{\gamma})'(t)=u'(t)+iv'(t)=0\Longrightarrow{u'(t)=0=v'(t)}[/itex], which means that [itex]u(t)=C_1, v(t)=C_2[/itex] are constant functions and so [itex](f\circ{\gamma})(t)[/itex] is constant.

    Must I use path integrals here. Is the above argumentation incorrect or imprecise?
    Last edited: Oct 21, 2007
  5. Oct 21, 2007 #4


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    Ok, now it looks good and complete!

    (But with path integrals it is very fast: "Let a,b be in U and y be a path in U joining and and B. Then by the FTC for path integrals in the complex plane, the integral of f '(z) along y is both 0 and f(b)-f(a). Therefor f(b)-f(a)=0. QED)
  6. Oct 21, 2007 #5
    Ok, I'll try to go into detail, because I'd like to write it down formally and clearly.

    So you say that:

    Constant functions are trivially continuous and holomorphic everywhere in C. U is open, so f' is holomorphic on U. Also the path gamma is continuous and hence smooth. These two statements imply that a primitive of f' on U exists and is determined up to a constant. Clearly, f' has a primitive on gamma.

    Let F be a primitive. Again, differentiability of f implies that primitive F is holomorphic and F=f.

    Now we apply the FTC to any path gamma between any two points a and b to conclude that:
    [itex]\displaystyle\int^{}_{\gamma} f'(z)\,dz = f(b)-f(a)[/itex]
    At the same time: [itex]\displaystyle\int^{}_{\gamma} f'(z)\,dz = 0[/itex] by hypothesis. Hence f(b)=f(a) for all different a and b in U. Thus it follows f is constant.

    Is it OK?

    Many thanks in advance
    Last edited: Oct 21, 2007
  7. Oct 21, 2007 #6


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