uniform convergence of $f_n=z^2-\frac{1}{n}$

Question

I have been trying to solve this for few days.

But dont know how to solve as I dont know how to use the equation

$$|f_n(z)-f(z)|<\epsilon$$

I mean how to select the $f(z)$ so as to solve for uniform convergence... Can anyone do this for me so i can review the process. I see some solved examples and completely understood them but as those were not unsolved so didnt get the idea how to do these tough ones.

Answer 1

The sequence is converging uniformly (I will highlight statements such as there exists or for all, so you see exactly about what numbers I am talking):

Definition. A sequence of functions over $\Bbb C$ taking complex values $(f_n)_{n\in\Bbb N}$ converges to a function $f:\Bbb C\to \Bbb C$ on $\Bbb C$ if and only if $$\forall \varepsilon>0: \exists N\in\Bbb N: \forall z\in\Bbb C, n\geq N: |f_n(z)-f(z)|<\varepsilon.$$

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(i) Claim. The $f_n$ converge uniformly to $f:\Bbb C\to\Bbb C:z\mapsto z^2$.

Proof. In our case, we have $|f_n(z)-f(z)|=|z^2-\frac1n-z^2|=|-\frac1n|=\frac1n$ for all $z\in\Bbb C$ and $n\in\Bbb N$. This makes the proof very easy: Given any $\varepsilon>0$, we note that for all complex numbers $z$: $|f_n(z)-f(z)|=\frac1n<\varepsilon$ for all $n>\frac1\varepsilon$. By the archimedean property, there exists a natural number $N>\frac1\varepsilon$. This achieves a proof.

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(ii) Note that $f_n^2(z)=z^4-\frac2n z^2+\frac1{n^2}$ for all $z\in\Bbb C$.

Claim. The $f_n$ don't converge uniformly to any function on $\Bbb C$.

Proof. Check that $f_n(z)\xrightarrow{n\to\infty} f(z)$ for all complex $z$; where $f:\Bbb C\to\Bbb C:z\mapsto z^4$. Note that the uniform limit of the $f_n^2$ is, if it exists, unique. And also, if the uniform limit exists, it must coincide with the pointwise limit. We thus only have to check if the $f_n$ converge uniformly to $f$. This is not the case:

Note that $|f_n(z)-f(z)|=|-\frac2nz^2+\frac{1}{n^2}|$ for all complex $z$. However, the latter expression equals $|-2n+\frac1{n^2}|$ whenever $z=n$. This is cleary not smaller than any $\varepsilon>0$ no matter how big $n$ is. Thus, there is no uniform convergence.

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(iii) Claim. The $f_n^2$ converge to the same $f$ as in (ii) on the unit circle, but this time also uniformly.

Proof. Left as an exercise to you.

Hint. $|-\frac2nz^2+\frac{1}{n^2}|\le\frac1{n^2}+\frac2n|z^2|=\frac1{n^2}+\frac2n|z|^2\le\frac1{n^2}+\frac2n$ for all natural numbers $n$, and $z$ in the complex unit disk.