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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.19(a),7.20(a),7.21

What are the errors, if any, in the following proofs? I put $\color{blue}{\text{highlights of proofs in blue}}$.

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  1. Exer 7.19 (a)

Obviously $f$ converges pointwise to zero. For uniform: Given $\varepsilon$, we must find $N$ that is independent of $z$ for convergence to be uniform convergence. We're given that for the same $\varepsilon$, $\varepsilon > |a_n|$ for $n > A$. $\therefore$, to have$|f_n(z)| \le a_n = |a_n| \le \varepsilon$, $\color{blue}{\text{I'll choose N=A}}$, independent of $z$.

  1. Exer 7.20 (a)

We're given that $$\forall \varepsilon > 0, \exists N > 0: \forall z \in G, n > N \implies |f_n(z)| < \varepsilon.$$ We must show that $$\forall \ \text{the same} \ \varepsilon > 0, \exists N_1 > 0 : n > N_1 \implies |f_n(z_n)| < \varepsilon$$

I think we can $\color{blue}{\text{choose z to be the individual elements}}$ $z_n$ of the sequence with $N=N_1$.

  1. Exer 7.21

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$f_n$ is cont so $f_n \stackrel{u}{\to} f \implies \color{blue}{\text{f is cont by Prop 7.25 (*)}}$ for $G = [0,\pi]$ ↯

  1. What's the connection between Exer 7.19(a) and the Weierstrass M-Test (**) ? They both seem to conclude uniform convergence from being bounded by something convergent. Is it really that they are analogues but that neither implies the other? Or could some additional condition with 1 of them imply the other? I guess Exer 7.19(a) wouldn't imply the second part of the Weierstrass M-Test namely that $\sum_n |f_n|$ converges uniformly.

(*) Prop 7.25

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(**) Weierstrass M-Test 7.28

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    $\begingroup$ 7.20(a) : just say : as $\left|fn(z)\right|<\epsilon$ for any $z$, it has to be true for $z=z_n$ (specialization). $\endgroup$ – Nicolas FRANCOIS Aug 2 '18 at 11:37
  • $\begingroup$ @NicolasFRANCOIS Thanks! Post as answer? 7.19(a) Does $N=A$ work? 7.21 is this right? $\endgroup$ – BCLC Aug 2 '18 at 11:40
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    $\begingroup$ Yup for 7.19, although I wouldn't say it that way, and yup for 7.21, just add "contradiction" ;-) $\endgroup$ – Nicolas FRANCOIS Aug 2 '18 at 11:45
  • $\begingroup$ @NicolasFRANCOIS You mean 'Yup for 7.19' ? Thanks! $\endgroup$ – BCLC Aug 2 '18 at 11:45
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    $\begingroup$ What exactly do you mean ? You want to apply Weierstrass condition to find another proof of 7.19(a) ? $\endgroup$ – Nicolas FRANCOIS Aug 3 '18 at 17:24
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For 7.19(a) implying Weierstrass, you have to remember that in general, studying a series implies studying its remainder. So let $$S_n=\sum_{k=0}^N f_k\qquad \text{and}\qquad R_n=\sum_{k=N+1}^\infty f_k$$ The first point in the demonstration of proposition 7.28 tells you that the series $\sum f_n(z)$ converges (absolutely) for any $z$, so $R_N$ is defined on $G$.

Now you can apply 7.19(a) by putting $a_N=\sum_{k=N+1}^\infty M_k$, which by hypothesis has limit $0$, to prove that $(R_n)$ converges uniformly to the null function, which is equivalent to saying that $\sum f_n$ converges uniformly.

The same argument applies if you replace $f_n$ by $\left|f_n\right|$.

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  • $\begingroup$ Thanks! Wait...sooo Exer7.19(a) DOES imply (both the fn and |fn| parts of) the Weierstrass M-test? $\endgroup$ – BCLC Aug 4 '18 at 11:12
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    $\begingroup$ Yup. The reciprocal is not so easy, because generally for studying sequences using series, you let $g_n=f_n-f_{n-1}$, but I can't figure out a way to translate the inequalities... Maybe it's vacation time :-) $\endgroup$ – Nicolas FRANCOIS Aug 4 '18 at 11:14
  • $\begingroup$ I don't think Weierstrass implies 7.19(a) because usually the sequences stuff implies the series stuff iirc. Right? Idk lol thanks!! $\endgroup$ – BCLC Aug 4 '18 at 11:38

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