# Uniform convergence of $\frac{1}{1+nz}$, and do we have 'piecewise' uniform convergence?

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) $\frac{1}{1+nz}$

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 $K$: $N=\sum_{k \in K}N_k 1_{z \in A_k}$.

• 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 $0$. – Ovi Aug 4 '18 at 2:15
• @Ovi lol I was overthinking I guess. Thanks! That settles the pointwise convergence I guess. Then there's the uniform convergence – BCLC Aug 4 '18 at 4:00
• @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 $|1/(1+nz)|\leq 1/(n|z|-1).$ – DanielWainfleet Aug 6 '18 at 3:25

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 $n\geqslant1$.