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

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

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


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

  • $\begingroup$ Your answer is not correct. $\endgroup$ Oct 31 '19 at 19:52

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


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


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


$$|f_n(x_n)-f(x_n)|\le \sup_{(0,1)}|f_n-f|$$ 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 $0<a<1$.

  • $\begingroup$ I dont understand why your taking a sequence $x_n$? How did you arrive at n to that power? $\endgroup$
    – John
    Oct 31 '19 at 19:27
  • $\begingroup$ @Amir To make the term $n^3x^2$ in denominator constant. $\endgroup$ Oct 31 '19 at 19:36
  • $\begingroup$ Why are you suggesting substituting a sequence into the expression i am not following that? $\endgroup$
    – John
    Oct 31 '19 at 19:36
  • $\begingroup$ @Amir It is a well known technic. For example if you have a term such $ n^2x+3$ you will consider $x_n=\frac{1}{n^2}$. $\endgroup$ Oct 31 '19 at 19:48
  • $\begingroup$ 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 $\endgroup$
    – John
    Oct 31 '19 at 19:57

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