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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 $(0,1)$.

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Hint:

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

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  • $\begingroup$ How did you decide to pick $x_n = \frac{1 - 1/n}{1 + 1/n}$ ? $\endgroup$ – ted Jul 13 '18 at 13:48
  • $\begingroup$ @ted: One could find the exact maximum or simply find a sequence where $|f_n(x_n)| \not\to 0$ as I am doing. You suspect the behavior near $x = 1$ prevents uniform convergence. The sequence $(1-1/n)$ is convenient to demonstrate non-uniform convergence of $x^n$ since $(1- 1/n)^{n} \to e^{-1} \neq 0$. Expressing as $f_n(x) = n^{3/2}x^{n-1}(1 - (1+1/n)x)$ we see that setting $x_n = (1-1/n)/(1+1/n)$ we still have $x_n \to 1$ but the expression in parentheses simplifies to $1/n$ leading to the last expression on the RHS in my hint. It is then easy to show $f_n(x_n) \to \infty$. $\endgroup$ – RRL Jul 13 '18 at 18:18
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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.

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