I have a practice question on uniform convergence and pointwise functions which has me confused.
The function is:
$ f_n : [0, M]\rightarrow\Bbb{R}$, $f_n(x)=\frac{(n+1)(2n+1)}{6n^2}x^3 $ for fixed and finite $M>0$
I'm pretty weak on how to handle a problem like this. As I understand it the pointwise function $f(x)$ in this case is found by taking $\lim_{n\to\infty}$ $f(x)=\frac{1}{3}x^3$, but I'm confused about how to then prove this (uniformly) converges.
Any suggestions as to where I'm going wrong or an explanation about how to prove uniform convergence would be greatly appreciated.
$|\frac {(n+1)(2n+1)} {6n^{2}}x^{3}-\frac 1 3 x^{3}|\leq M^{3} |\frac {(n+1)(2n+1)} {6n^{2}}-\frac 1 3 | $ for all $x \in [0,M]$ so $\sup \{|f_n(x)-f(x)|: 0 \leq x \leq M\} \leq \frac {(n+1)(2n+1)} {6n^{2}}-\frac 1 3$ and and $\frac {(n+1)(2n+1)} {6n^{2}}-\frac 1 3 \to 0$. Hence, the convergence is uniform.