# Checking the uniform convergence of sequence of functions

I have been trying some questions on uniform convergence.Got stuck in one of those questions which says that

For a positive real number p, let (f$$_n$$) is a sequence of functions defined on [$$0,1$$] by

$$f_n(x) = \begin{cases} n^{p+1}x, \text{if 0 \le x \lt \frac{1}{n}}\\ \frac{1}{x^p}, \text{if \frac{1}{n} \le x \le 1} \end{cases}$$

I have found its point-wise limit given by

$$f(x)= \begin{cases} 0, \text{if x = 0}\\ \frac{1}{x^p}, \text{if 0 \lt x \le 1} \end{cases}$$

I am stuck in proving whether its uniformly convergent or not.

I take any $$\epsilon$$ $$\gt$$ $$0$$.Now I need to know that does there exist a natural number m such that $$\lvert f_n(x)-f(x)\rvert$$ $$\lt$$ $$\epsilon$$ for all n $$\geq$$m and for all $$x$$ in [$$0,1$$]?

$$|f_n(\frac 1 {n^{p+1}}) -\frac 1 {(\frac 1 {n^{p+1}})^{p}}|=|1-n^{p(p+1)}| \to \infty$$. Hence $$\sup_x |f_n(x)-f(x)|$$ does not tend to $$0$$ and the convergence is not uniform
• Just for variety, and no better than the answer given, another approach is to observe that each $f_n$ is continuous on $[0,1]$ while $f$ is not. Uniform convergence of continuous functions has a continuous limit (if that theorem is available to you), so the convergence cannot be uniform. Jun 4, 2020 at 14:23