Uniform convergence $f_n(a) = a^{4n} + \frac1{n^2}$

Question

I have $f_n(a) = a^{4n} + \frac1{n^2}$ which I know converges to $f(a)=0$ uniform on theinterval $[0,1)$

This works?

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$\lim \limits_{n \to \infty} a^{4n} + \frac1{n^2} = \lim \limits_{n\to\infty} a^{4n} + \lim \limits_{n\to\infty} \frac1{n^2} = \lim \limits_{n\to\infty} a^{4n} + 0 = \lim \limits_{n\to\infty} a^{4n}$

$\lim \limits_{n\to\infty} a^{4n} = 0$

Proof: $\forall a \in [0,1), \epsilon \gt 0, \exists N\in\mathbb{N}$ such that $n\gt N$

$a^{4n} \lt a^{4N}$

Set $\epsilon = a^{4N}$

$a^{4n} \lt \epsilon$, therefore $\forall a\in [0,1) $, $\lim \limits_{n\to\infty} a^{4n} = 0 $

$f_n(a) = a^{4n} + \frac1{n^2}$ converges uniform on $[0,1)$.

Answer 1

I am afraid your proof is not sufficient for uniform convergence. What you show is only pointwise convergence.

$f_n\to f$ pointwise $\iff \forall a\in[0,1),\forall\epsilon>0,\exists N=N(a,\epsilon)\in\mathbb{N},\forall n\geq N: |f_n(a)-f(a)|<\epsilon$

$f_n\to f$ uniformly $\iff \forall\epsilon>0,\exists N=N(\epsilon)\in\mathbb{N},\forall n\geq N,\forall a\in[0,1): |f_n(a)-f(a)|<\epsilon$

For pointwise convergence you do start by picking a point on the domain, so that the index $N$ you come up with depends on that $a$ (by the simple virtue that you first pick a point), while for uniform convergence the index $N$ you chose works for any point on the domain. Intuitively uniform convergence is thus stronger than pointwise convergence (formally too, of course).

In your proof you first pick a point, pick $\epsilon$ and $N$ accordingly, and then pick yet another $a$.

Also, are you sure that $f_n\to 0$ uniformly? How about $\lim f_n(x_n)$, where $x_n:=\left(\dfrac{1}{2}\right)^{1/n}$?

Answer 2

Consider the function $g(a)=|f_{n}(a)-f(a)|=f_{n}(a)-f(a)=a^{4n}+\frac{1}{n^{2}}$ in the interval $[0,1)$.

Now find the maximum value of $g$ in $[0,1)$.

$g$ attains its maximum value when $a\to 1$.

When $a\to 1$ then $g(a)\to 1+\frac{1}{n^{2}}\to 1$ as $n\to \infty$.

So , $sup\{|f_{n}(a)-f(a)|:a\in [0,1)\}\not \to 0$ as $n\to \infty.$

Therefore $f$ is not uniformly convergent in $[0,1)$.