# Evaluating Convergence (Uniform)

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:-

$$g_n:(0,1)\to \mathbb{R}$$ $$g_n(x) = \frac{n^3+1}{n^3x^2+1}, x\in(0,1)$$

Checking point wise convergence first

$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{n^3+1}{n^3x^2+1}$$

Dividing by $$n^3$$gives the following :-

$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{1+\frac{1}{n^3}}{x^2+\frac{1}{n^3}}$$

Taking the Limit as n $$\to \infty$$ gives the following

$$\lim_{n\to \infty}g_n(x) = \frac{1+\frac{1}{\infty^3}}{x^2+\frac{1}{\infty^3}}$$

$$\lim_{n\to \infty}g_n(x) = \frac{1+0}{x^2+1} = \frac{1}{x^2}$$

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

$$M_n = sup|f_n(x)-f(x)|,x\in \mathbb{R}$$

$$|f_n(x)-f(x)|$$ $$|\frac{n^3+1}{n^3x^2+1} - \frac{1}{x^2}|$$

$$\frac{(n^3x^2+x^2)-(n^3x^2+1)}{(n^3x^2+1)(x^2)}$$

$$|\frac{x^2-1}{(n^3x^2+1)(x^2)}|$$ The mod gives $$\frac{x^2+1}{(n^3x^2+1)(x^2)}$$

is it accurate to say the following when checking to see uniform convergence

$$\frac{x^2+1}{(n^3x^2+1)(x^2)} < \frac{1}{n^3}$$

$$\lim{n \to \infty}$$

Therefore I can conlclude that $$sup|f_n(x)-f(x)|\to 0$$

Therefore the function is uniformly convergent? Oh am i wrong in my evaluation?

• Your answer is not correct. – hamam_Abdallah Oct 31 '19 at 19:52

For $$x\in (0,1)$$ and $$n$$ large enough,

$$g_n(x)=|f_n(x)-\frac{1}{x^2}|=\frac{|x^2-1|}{(n^3x^2+1)x^2}$$

Now take the sequence $$(x_n)$$ such that

$$n^3x_n^2=1$$ or

$$x_n=n^{-\frac 32}=\frac{1}{n^{\frac 32}}$$

Then, $$x_n\in (0,1)$$ and

$$|f_n(x_n)-f(x_n)|=\frac{|n^{-3}-1|}{2n^{-3}}$$

$$=\frac 12|1-n^{3}| \to +\infty$$

But

$$|f_n(x_n)-f(x_n)|\le \sup_{(0,1)}|f_n-f|$$ thus $$\lim_{n\to+\infty}\sup_{(0,1)}|f_n-f|=+\infty$$ The convergence is not uniform at $$(0,1)$$.

It is uniform at $$(a,1)$$ with $$0.

• I dont understand why your taking a sequence $x_n$? How did you arrive at n to that power? – Amir Oct 31 '19 at 19:27
• @Amir To make the term $n^3x^2$ in denominator constant. – hamam_Abdallah Oct 31 '19 at 19:36
• Why are you suggesting substituting a sequence into the expression i am not following that? – Amir Oct 31 '19 at 19:36
• @Amir It is a well known technic. For example if you have a term such $n^2x+3$ you will consider $x_n=\frac{1}{n^2}$. – hamam_Abdallah Oct 31 '19 at 19:48
• yeah i realised but i am not familiar just started evaluating question like these i dont understand well im not getting $2n^-3$ im confused about that i got the top term – Amir Oct 31 '19 at 19:57