Hi Guys was attempting this question and was wondering if I was doing the question correctly?
Determine whether or not the sequence of functions is uniformly convergent:-
$$g_n:(0,1)\to \mathbb{R}$$ $$g_n(x) = \frac{n^3+1}{n^3(x^2+1)}, x\in(0,1)$$
Checking point wise convergence first
$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{n^3+1}{n^3(x^2+1)}$$
Dividing by $n^3$gives the following :-
$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{1+\frac{1}{n^3}}{x^2+1}$$
Taking the Limit as n $\to \infty$ gives the following
$$\lim_{n\to \infty}g_n(x) = \frac{1+\frac{1}{\infty^3}}{x^2+1}$$
$$\lim_{n\to \infty}g_n(x) = \frac{1+0}{x^2+1} = \frac{1}{x^2+1}$$
$$\lim_{n\to \infty}g_n(x) = \frac{1}{x^2+1} = f(x)$$
Therefore by point wise convergence the sequence of functions converges to the previous function.
In order to determine the uniform convergence we must analyze the follwing
$$M_n = sup|f_n(x)-f(x)|,x\in \mathbb{R}$$
But in this question we are dealing with an open interval so I am assuming that this is not uniformly convergent based on the fact for an open interval the supremum doesn't exist.
I do know that the maximum value of the function will be a value or number as close to 1 but that number can be any number and so i am trying to understand how to conclude this question.
$$g_n:(0,1)\to \mathbb{R}$$
$$f(x) = \frac{1}{x^2+1}$$
so just to evaluate to get an idea of the smallest value and the largest value but considering these value are defined in an open interval
$$f(0) = \frac{1}{0+1}=1 $$ $$f(1) = \frac{1}{1+1}=\frac{1}{2} $$
therefore since 1 is the largest value but its an open interval i am seeing that value will get as close to a 1/2 but will never reach that value and for the other situation values will get as close to 1 but never reach that value can anyone shed some light as how to deal with this open interval as it relates to analyzing the question to see if its uniformly convergent ?
If $x\in(0,1)$ and $n\in\mathbb N$, then$$\bigl\lvert g_n(x)-g(x)\bigr\rvert=\frac{1/n^3}{1+x^2}<\frac1{n^3},$$since $1+x^2>1$. So, $\sup_{x\in(0,1)}\bigl\lvert g_n(x)-g(x)\bigr\rvert\leqslant\frac1{n^3}$ an therefore, since $\lim_{n\to\infty}\frac1{n^3}=0$, we have$$\lim_{n\to\infty}\sup_{x\in(0,1)}\bigl\lvert g_n(x)-g(x)\bigr\rvert=0,$$and so the convergence is uniform.