How to show the sequence of functions $f_n(x) = \sqrt{n}(nx^{n-1} - (n+1)x^n)$ not uniformly convergent

Question

I'm having trouble showing that the sequence of functions $\{ f_n \}_{n=1}^\infty$ is not uniformly convergent, where $$ f_n(x) = \sqrt{n}(nx^{n-1} - (n+1)x^n) $$ and $x \in (0,1)$.

Perhaps one way to show this is that the function $f(x) \triangleq \lim_{n\to\infty} f_n(x)$ is discontinuous if we extend the domains of the functions $\{ f_n \}$ to $(0,1]$.

Here we would have $$ f(x) = \begin{cases} 0 & \text{ if } x \in (0,1)\\ 1 & \text{ if x = 1. } \end{cases} $$ Since $f(x)$ is not continuous on $(0,1]$, the functions $\{ f_n \}$ cannot converge uniformly to $f(x)$.

But I'm not sure this reasoning applies if the functions $f$ and $\{ f_n \}$ are only defined on the open interval $(0,1)$.

Answer 1

Hint:

With $\displaystyle x_n = \frac{1 - 1/n}{1 + 1/n} \in (0,1)$, we have $$\sup_{x \in (0,1)} |f_n(x)| \geqslant |f_n(x_n)| =n^{3/2}\frac{(1 - 1/n)^{n-1}}{(1 + 1/n)^{n-1}}\left(1 - (1 + 1/n)\frac{1 - 1/n}{1 + 1/n} \right) \\ = n^{1/2}\frac{(1 - 1/n)^{n-1}}{(1 + 1/n)^{n-1}} $$

Answer 2

Your suggested proof fails for the reason you give. Here's a hint: since we know $f_n$ converges to $0$ pointwise on $(0,1),$ if the convergence is uniform then $M_n\to 0$ where $M_n$ is $\sup_{x\in(0,1)}|f_n(x)|$ Furthermore, this condition is sufficient for uniform convergence, so check out the maxima.