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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.22

Where do the following sequences converge pointwise? Do they converge uniformly on this domain? (a)$\frac{nz^n}{1}$ (b) $\frac{z^n}{n}$ (c) $ \frac{1}{1+nz}$

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These sequences were asked about in these questions: Uniform convergence of $\frac{1}{1+nz}$, Where do the following sequences converge pointwise and uniformly? and $\sum_{n=1}^\infty\frac{z^n}{n}$ does not converge uniformly on $\mathbb{D}$. , but I have other questions. ((a) and (b) are here)

Firstly, is this supposed to be unambiguous? I mean 'this domain' refer to the answers we got for pointwise convergence? Or are 7.22(a) and (b) missing specified 'domains', which I guess would be what (c) has? Anyhoo, I'm just gonna assume 'this domain' refer to the answers we got for pointwise convergence for (c). I'm gonna try with the $\Re(z) \ge 0$ for (c2), (c3), (c7) and (c8) and then taking off $\Re(z) \ge 0$ in (c4), (c5), (c9) and (c10). Finally in relation to question (c), I have questions (d1)-(d4).

(c1)

Pointwise convergence: $\lim \frac{1}{1+nz} = 1 \iff z=0?$

$$\lim \frac{1}{1+nz} = \lim \frac{1}{1+n(0)} = \lim 1 = 1$$

(c2)

Pointwise convergence: $\lim \frac{1}{1+nz} = 0 \iff |z| \ge r > 0, \Re(z) \ge 0, z \ne 0?$

$$\lim \frac{1}{|1+nz|} \stackrel{\Re(z) \ge 0}{\le} \lim \frac{1}{|nz|} \le \lim \frac{1}{n|z|} \le \lim \frac{1}{nr} = 0$$

Where do I use $z \ne 0$, whether or not this is wrong? If wrong, then which part and why?

(c3)

Pointwise convergence: $\lim \frac{1}{1+nz} = 0 \iff \Re(z) \ge 0, z \ne 0?$

Deduced from previous (c2)? Equivalent to previous actually?

(c4)

Pointwise convergence: (I'm taking off $\Re(z) \ge 0$ in (c2))

$\lim \frac{1}{1+nz} = 0 \iff |z| \ge r > 0, z \ne 0?$

$$\lim |\frac{1}{1+nz}| = \lim \frac{1}{|1+nz|} \le \lim \frac{1}{|1-n|z||} \le \lim \frac{1}{|1-nr|} = 0$$

Where do I use $z \ne 0$, whether or not this is wrong? If wrong, then which part and why?

(c5)

Pointwise convergence: (I'm taking off $\Re(z) \ge 0$ in (c3))

$\lim \frac{1}{1+nz} = 0 \iff z \ne 0?$

Deduced from previous (c4)? Equivalent to previous actually?

(c6)

Uniform convergence: $\lim \frac{1}{1+nz} \stackrel{u}{=} 1 \iff z=0$ trivially?

(c7)

Uniform convergence: $\lim \frac{1}{1+nz} \stackrel{u}{=} 0 \iff |z| \ge r > 0, \Re(z) \ge 0, z \ne 0?$

Bounded by a convergent sequence

(c8)

Uniform convergence: $\lim \frac{1}{1+nz} \stackrel{u}{=} 0 \iff \Re(z) \ge 0, z \ne 0?$

Also deduced from previous (c7)? Equivalent to previous actually?

(c9)

Uniform convergence: (I'm taking off $\Re(z) \ge 0$ in (c7))

$\lim \frac{1}{1+nz} \stackrel{u}{=} 0 \iff |z| \ge r > 0, z \ne 0?$

Bounded by a convergent sequence

(c10)

Uniform convergence: (I'm taking off $\Re(z) \ge 0$ in (c8))

$\lim \frac{1}{1+nz} \stackrel{u}{=} 0 \iff z \ne 0?$

Also deduced from previous (c9)? Equivalent to previous actually?

(My own questions)

(d1)

Is it possible to have some function sequence uniformly convergent to $a \in \mathbb C$ on $A \subseteq \mathbb C$ and then $b \in \mathbb C$ on $B \subseteq \mathbb C$?

This arises from (c) where it seems like it's uniformly convergent to 1 at the point $z=0$ and then uniformly convergent to 0 at $\Re(z) \ge 0, z \ne =0$. This also arises from a mistake I made in (c) where I supposedly proved uniform convergence to 0 on $|z| \le 1$ (Then I wondered where I used $z \ne 0$ which led me to realise I was wrong).

(d2)

If yes to (d1) and in the case that $a \ne b$, then do we necessarily have that the function sequence is not uniformly convergent on $A \cup B$?

(d3)

If yes to (d1) and in that case that $a=b$, then is it possible that the function sequence is not uniformly convergent on $A \cup B$?

(d4)

Is there some kind of concept of 'piecewise' uniform convergence? I was thinking $N=N_1 1_{z \in A_1} + N_2 1_{z \in A_2}$ My understanding of uniform convergence that is $N$ is independent of $z$. I was thinking we define 'piecewise' for countable indices $K$: $N=\sum_{k \in K}N_k 1_{z \in A_k}$.

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    $\begingroup$ Hey, I don't know much about complex analysis but you put a qustion mark after $\lim \frac{1}{1+nz} = 1 \iff z=0$, so I tried to think about it; if you write $z = a+bi$ and multiply the numerator and denominator by the conjugate you get $\lim \dfrac {an+1}{(an+1)^2 - b^2} - \dfrac {b}{(an+1)^2 - b^2}i$ which more clearly goes to $0$. $\endgroup$ – Ovi Aug 4 '18 at 2:15
  • $\begingroup$ @Ovi lol I was overthinking I guess. Thanks! That settles the pointwise convergence I guess. Then there's the uniform convergence $\endgroup$ – BCLC Aug 4 '18 at 4:00
  • $\begingroup$ @Ovi. We have $|1/y|=1/|y|$ for any non-zero $y\in \Bbb C.$ And we have $ |1+nz|\geq |nz|-|1|=n|z|-1.$ So if $z\ne 0$ then for all $n>1/|z|$ we have $|1/(1+nz)|\leq 1/(n|z|-1). $ $\endgroup$ – DanielWainfleet Aug 6 '18 at 3:25
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I've not read whole of your post, but apparently has useful remarks. For pointwise convergence of $f_n(z)=\dfrac{1}{1+nz}$, clearly for every $z_0\neq0$ and $\varepsilon>0$ we simply write $$\varepsilon>|f_n(z_0)-0|=\dfrac{1}{|1+nz_0|}\geqslant\dfrac{1}{1+n|z_0|}\geqslant\dfrac{1}{n(1+|z_0|)}$$ then it is sufficient to let $N(\varepsilon,z_0)\geqslant\lfloor\dfrac{1}{\varepsilon(1+|z_0|)}\rfloor$. Therefore $f_n(z)$ converges pointwise to $$ f(z)= \begin{cases} 1&z=0,\\ 0&z\neq0. \end{cases} $$ and this convergence is not uniform, consider $z_n=\dfrac1n$ so $$|f_n(z_n)-f(z_n)|=\left|f_n(\dfrac1n)-f(\dfrac1n)\right|=\dfrac12$$ then this convergence isn't uniform for $|z|\leqslant2$, but for $|z|\geqslant2$, then $$|f_n(z)-0|=\dfrac{1}{|1+nz|}\leqslant\dfrac{1}{n|z|-1}\leqslant\dfrac{1}{2n-1}\leqslant\dfrac{1}{n}<\varepsilon$$ for all $n\geqslant1$.

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  • $\begingroup$ Thanks user 108128. I'll analyse later. $\endgroup$ – BCLC Aug 5 '18 at 12:02

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