5

I'm trying to prove that the complex logarithm function is continuous using this theorem, but I'm hitting a snag in part of the proof.

Let $D=\Bbb C\setminus(-\infty,0]$. The claim is that the function $\log\upharpoonright D:D\to\Bbb C$ is continuous on $D$. What we know so far is that $\log$ is defined as the inverse of $\exp\upharpoonright (\Bbb R\times(-\pi,\pi])$ (where the notation refers to the set of all complex numbers with real part in $\Bbb R$ and imaginary part in $(-\pi,\pi]$), and it is well-defined because we have shown that $\exp$ on this domain is a bijection onto $\Bbb C\setminus\{0\}$. We also know that $\exp$ is continuous.

Given $x\in D$, we wish to show that $\log$ is continuous at $x$. In order to apply the linked theorem, we need a compact region, so let $y=\log x$ and define $Y=[\Re y-1,\Re y+1]\times[\frac{\Im y-\pi}2,\pi]$. Then $Y$ is compact, so $\exp(Y)$ is also compact, and since $y\in \Bbb R\times(-\pi,\pi]$ follows from the definition of $\log$ and $\Im y\ne\pi$ because this would imply $x\in(-\infty,0]$, we also have $$y\in Y^\circ=(\Re y-1,\Re y+1)\times(\frac{\Im y-\pi}2,\pi).$$

Now we can apply the theorem to deduce that $\exp\upharpoonright Y:Y\to\exp(Y)$ is a homeomorphism, so $\log$ is continuous on $\exp(Y)^\circ$. Where I got stuck is in the last part, to show that $x\in\exp(Y)^\circ$ given that we already know $x\in\exp(Y^\circ)$, because the subspace topologies involved don't play well with interior here. Specifically, we know that $\exp(Y^\circ)$ is open in $\exp(Y)$, but I don't see how this implies that it is open in $\Bbb C$ (or $D$).

I realize that I can probably grok this proof with sufficient details of the shape of the transformed region $\exp(Y)$, but I'm going for maximum "slick"-factor with this proof as well, so I'd prefer to avoid any calculations more complicated than necessary. In particular, if possible I don't want to use any other properties of the exponential function than those mentioned here.

If there is another entirely different way to prove this nicely, I'm all ears.

Community
  • 1
Mario Carneiro
  • 27,438
  • How about using the power series expansion and the fact that the uniform limit of continuous functions is continuous? – Alp Uzman Feb 18 '15 at 09:28
  • @Uzman Of $\exp$ or of $\log$? If you mean $\log$, I don't have it. All I have is the definition as an inverse function to $\exp$. I hope to follow this theorem up with a proof that $\log'=1/x$ (which I'm also not sure about) - if I had that, I would already be done. – Mario Carneiro Feb 18 '15 at 09:30

1 Answers1

1

By definition, and upon choosing a branch cut ( say, $\;\{z\in\Bbb C\;:\;\;\text{Re}\,z\le 0\}$ ) , we have that

$$z=x+iy\;,\;\;x,y\in\Bbb R\;\implies\;\text{Log}\,z:=\log|z|+i\arg z=\frac12\log(x^2+y^2)+i\arctan\frac yx$$

Putting $\;u(x,y)=\frac12\log(x^2+y^2)\;,\;\;v(x,y):=\arctan\frac yx\;$ , we get:

$$\begin{align}&u_x'=\frac x{x^2+y^2}\;,\;\;v_y'=\frac1x\frac1{1+\frac{y^2}{x^2}}=\frac x{x^2+y^2}\\{}\\ &u_y'=\frac y{x^2+y^2}\;,\;\;v_x'=-\frac y{x^2}\frac1{\frac{y^2}{x^2}}=-\frac y{x^2+y^2}\end{align}$$

We can see our function fulfills the Cauchy- Riemann Equations . This, together with the fact that each partial derivative is a continuous function in the chosen domain, makes Log$\,z\;$ an analytic function there and, thus, continuous.

This does not use what you mentioned at the beginning of your question, but it is something pretty different as you ask at the end of it.

Timbuc
  • 34,191
  • This would be an awesome proof if I could use it, but it uses the derivative of log as an (apparently) essential ingredient, which seems to be begging the question. – Mario Carneiro Feb 18 '15 at 10:05
  • Well, you use the derivative of the real logarithm function, which I'd say should be old, well-known stuff in complex variable. – Timbuc Feb 18 '15 at 10:12
  • Do you know a proof of the claim on page 6 of your link: (Inverse functions) If $f(z)$ is complex differentiable and one-to-one, with nonzero derivative, then the inverse function $f^{-1}(z)$ is also differentiable, and $$(f^{-1}(z))'=1/f'(f^{-1}(z)).$$ This would also do the trick, since I know the derivative of $\exp$. – Mario Carneiro Feb 18 '15 at 10:14
  • I'm working sort of out of order w.r.t the usual teaching sequence since this is a formal proof so I have to start at the logical bottom, not the conceptual bottom. I had intended derivative of real log to be a corollary of the complex proof. – Mario Carneiro Feb 18 '15 at 10:16
  • @MarioCarneiro As far as I am aware, the proof shall be practically identical to the one in the real case: Putting $;f(w)=z;,;;f(w_0)=z_0;$ , we get$$(f^{-1}(z_0))'=\lim_{z\to z_0}\frac{f^{-1}(z)-f^{-1}(z_0)}{z-z_0}=\lim_{z\to z_0}\frac{w-w_0}{f(w)-f(w_0)}=\frac1{f'(w_0)}$$ and etc. – Timbuc Feb 18 '15 at 10:20
  • @MarioCarneiro I think I understand what you say about the conceptual bottom, yet I don't understand so much why: if you already were going to use the approach you mention in that link you put, clearly basic topology and real analysis should be understood. The derivative of the inverse function's theorem is, as you can see above, very easily proved, so I don't think that going all the way to complex variable via topology is going to the logical bottom – Timbuc Feb 18 '15 at 10:24
  • The problem is that the proofs tend to have weird dependencies, so that "proved so far" touches on a wide range of difficulty levels at once. I am actually working on basic real analysis here, but my functions are defined on complexes so I guess I need to do some complex analysis. There are a lot of topological theorems available, so it makes that approach more convenient. As for your proof, is continuity of $f^{-1}$ already assumed there? – Mario Carneiro Feb 18 '15 at 10:32
  • @MarioCarneiro I can't understand how come you're working on basic real analysis yet your functions are defined on the complex field...as for the last part: of course, as it follows from the necessary assumption that $;f^{-1};$ is differentiable, otherwise we cannot move. – Timbuc Feb 18 '15 at 10:35
  • Note that the linked PDF claims that differentiability is provable, you only need to know that the original function is differentiable with nonzero derivative (in some ways that's the most important part of the claim). My functions are defined on $\Bbb C$ because I can't define a function twice - that's not logically sound - so I define it in the most general reasonable context. – Mario Carneiro Feb 18 '15 at 10:36
  • @MarioCarneiro Yes, apparently they mean that thin because of the proof as I wrote it above one can see that the limit exists by basic arithmetic of limits and because we're given $;f'\neq 0;$ in that point, so that $;(f^{-1})';$ exists – Timbuc Feb 18 '15 at 10:39
  • I don't understand what you mean by "I can't define a function twice". You can define $;\sin x;$ either for $;x\in\Bbb R;$ (basic trigonoemtry) or for $;x\in\Bbb C;$, say by means of the complex exponential function. – Timbuc Feb 18 '15 at 10:41
  • Regarding the claim that $(f^{-1})'$ exists, when I work out the difference quotient I get $(f^{-1})'(x)=\lim_{y\to x}\frac{f^{-1}(y)-f^{-1}(x)}{f(f^{-1}(y))-f(f^{-1}(x))}$ which looks like 1 over the diff. quotient of $f$, but only if we assume that $f^{-1}(y)\to f^{-1}(x)$ as $y\to x$, i.e. $f^{-1}$ is continuous. Once a function is defined, it goes into a database of true things. After that, you can't define it again to be something else, because that would be incorrect. $\operatorname{dom}\sin$ is a well-formed expression, so it needs to evaluate to exactly one thing, in my case $\Bbb C$. – Mario Carneiro Feb 18 '15 at 10:49
  • 1
    @MarioCarneiro Nop. Look at my shortened semiproof with that notation there: begin by assuming $;w\to w_0;$ in order to calculate the limit for $;\frac1{f'(w_0)};$ . Then, by continuity of $;f;$ , we get that also $;z=f(w)\to f(w_0)=z_0;$, which we'll use for the derivative of the inverse function, i.e.: read what I wrote from right to left – Timbuc Feb 18 '15 at 11:02