Prove $f_n(x)=\frac{x^n}{\sqrt{3n}}$ for $x \in [0,1]$ is uniformly convergent

Question

I'm completely confused by uniform convergence, but I put together the following proof just based on my other questions here and examples I read online.

Discussion: Let $\epsilon \gt 0$

We want to find $N$ so that for all $x$ in $[0,1]$ and $n \gt N$, $\left|f_n(x)-f(x)\right|=\left|\frac{x^n}{\sqrt{3n}}-0\right|\lt \epsilon$

$\left|\frac{x^n}{\sqrt{3n}}\right|\le\frac{1}{\sqrt{3n}}\le\frac{1}{\sqrt{3N}}\lt\epsilon$ if $\frac{1}{3\epsilon^2}\lt N$

Proof:

Give $\epsilon \gt 0$, let $N=\frac{1}{3\epsilon^2}$. Then for all $x$ in $[0,1]$ and $n \gt N$, $\left|f_n(x)-f(x)\right|=\left|\frac{x^n}{\sqrt{3n}}-0\right|\le\frac{1}{\sqrt{3n}}\le\frac{1}{\sqrt{3N}}=\frac{1}{\sqrt{\frac{1}{3\epsilon^2}}}=\epsilon$

I'm completely confused by this entire subject, I've been reading the definition over for the last 3 hours and I just do not understand what is going on.

Answer 1

Copper.hat has the answer, but you are confused, it seems, about the significance of uniformly convergent.

Try your same sort of proof on a different problem on $[0,1]$, namely, $$ f_n(x) = \frac{x^n}{(x-\frac12)^2\sqrt{3n}} $$ again with $f(x) = 0$ as the function they converge to.

In your original problem, if I give you $\epsilon$ you can come back to me with an $N$ such that whenever $n>N$, $|f_n(x)| < \epsilon$ for all $x \in [0,1]$.

In the new problem, for any given $x$ and $\epsilon$ you can find an $N$ such that whenever $n>N$, $|f_n(x)| < \epsilon$ for that $x$ -- but the $N$ you need to give me depends on the value of $x$ I gave you. This sequence of functions is therefore not uniformly convergent.

Answer 2

The uniform part means that for a given $\epsilon>0$ that there is some $N$ such that if $n \ge N$ then $|f(x)-f_n(x)| < \epsilon$ for all $x$.

If $x \in [0,1]$ then $|f_n(x)-0 | \le {1 \over \sqrt{3n}}$, so given $\epsilon>0$, choose any $N$ such that ${1 \over \sqrt{3N}} < \epsilon$.

Then, if $n \ge N$ we have ${1 \over \sqrt{3n}} \le {1 \over \sqrt{3N}} < \epsilon$ and so $|f_n(x)-0 | < \epsilon$ for all $x \in [0,1]$.

As an example of where this fails: Choose $g_n(x) = x^n$ for $x \in [0,1)$ (note the end point $1$ is excluded). Then for any fixed $x$ (in $[0,1)$) we have $g_n(x) \to 0$, but the convergence is not uniform. To see this, note that $g_n({1 \over \sqrt[n]{2}}) = {1 \over 2}$ for all $n$.