Let $(f_n)_n$ be a sequence of functions defined by $$f_n : [0,1] \rightarrow \mathbb{R}: x \mapsto nx^n(1-x). $$
We have $\lim_{n \to \infty} f_n(x) = 0$ for all $x$. My textbook says the convergence isn't uniformly though, and I don't understand why.
I computed $$\sup_{x \in [0,1]} |f_n(x) – 0 | = \sup_{x \in [0,1]} | nx^n (1-x) | = \ ? $$
How do I figure out this supremum?
When I use the definition, I see that $$| nx^n(1-x) – 0 | = | nx^n – nx^{n+1}| \leq | nx^n | + | nx^{n+1}| \leq n + n = 2n. $$ And I can never get this smaller than $\epsilon$. But I couldn't find an explicit $\epsilon > 0$ and a $x \in [0,1]$ such that $ |f_n(x)| \geq \epsilon. $
Any help is appreciated.
Best Answer
This is not a solution but it might be useful to others. I decided to investigate the functions geometrically, so I thought I would share in case it brings to light the methods which have been put forward by others.
This demonstrates (non-rigorously) that the function ought not to converge uniformly and it shows the maximum at $\frac{n}{n+1}$ approaching $\frac{1}{e}$ as $n$ grows.