$\lim \int_0^{\infty} \frac1n e^{-\frac{t}{n}} dt \ne \int_0^{\infty} \lim \frac1n e^{-\frac{t}{n}} dt$ in complex as in real

complex-analysiscontour-integrationreal-analysisuniform-convergence

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

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  1. I was about to say that Prop 7.27 doesn't apply because $\gamma = G$. Unfortunately, the inclusion is loose.

  2. Um, does $G$ have to be a region? If so, then I guess $\mathbb R_{\ge 0}$ is not a region? If not, then perhaps

  3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: \mathbb R \to \mathbb R$ ?

  4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: \mathbb R_{\ge 0} \to \mathbb C$ ?

  5. In re #3 and #4, what changes if we redefine $f_n: \mathbb R \to \mathbb C$ ?

Best Answer

We can extend the $\{f_n\}$ to functions $\tilde f_n:\mathbb C\to\mathbb R$ which converge uniformly on $\mathbb C$, by defining $\tilde f_n(z)=f_n(\text{Re}(z))$ for $Re(z)\geq0$ and $\tilde f_n(z)=\frac{1}{n}$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $\mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $\mathbb R_{>0}$, $\mathbb R$, $\mathbb C$, or any other reasonable subset of $\mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $\mathbb C$, and $\mathbb R_{\geq0}$ is not a compact subset of $\mathbb C$.

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