Does $\{(1-x)x^k\}$ converge uniformly on $[0,1]?$ The convergence is clearly pointwise to the function $f(x) = 0$, but I'm not sure if this convergence is uniform. I wanted to prove that it wasn't uniform because I had a feeling it wasn't, but I'm having trouble with the problem of finding $\varepsilon > 0$ such that $\forall N\in \mathbb N,$ $\exists x\in [0,1]$ such that
$$
(1-x)x^{N+1} \geq \varepsilon.
$$
Maybe the convergence is uniform after all? I tried proving that, but again, got stuck.
 A: If the uniform limit exists, then it is $f(x)=0$. The maximum of $f_n:=(1-x)x^n$ is at 
$$f'(x)=nx^{n-1} - (n+1)x^n=0 \Longrightarrow x=n/(n+1), $$
where 
$$f_n(n/n+1)=\frac1{n+1}\frac{n^n}{(n+1)^n}=\frac{n^n}{(n+1)^{n+1}}\to 0$$
as $n\to +\infty$.
A: A somewhat different argument, which is useful in related problems, is to notice that for each $\delta > 0$, uniform convergence on $[0,1-\delta]$ comes for free. Specifically, for $x \in [0,1-\delta]$, $(1-x)x^n \leq x^n \leq (1-\delta)^n$. For $x \in [1-\delta,1]$, you have $(1-x) x^n \leq 1-x \leq \delta$. So by taking $\delta<\varepsilon$ and then $n$ such that $(1-\delta)^n < \varepsilon$, you get the convergence you need. 
This idea can be used to prove that
$$\lim_{n \to \infty} \int_0^1 (n+1) x^n f(x) dx = f(1)$$
whenever $f$ is bounded on $[0,1]$ and continuous at $1$. The trick is to write
$$\int_0^1 (n+1) x^n f(x) dx - f(1) = \int_0^1 (n+1) x^n (f(x)-f(1)) dx$$
since $\int_0^1 (n+1) x^n dx = 1$. From there the argument is similar to the one above: on $[0,1-\delta]$ you get uniform convergence for free, and on $[1-\delta,1]$ you get an integrand bounded by $(n+1) x^n \varepsilon$, whose integral on $[1-\delta,1]$ is less than $\varepsilon$. Note that this shouldn't be too surprising, since this is an easy integration by parts problem when $f$ is $C^1$.
A: Hint: study where is the maximum of the positive function $f_k(x)=(1-x)x^k$ in $[0,1]$.
