I'm aware this question has been posted multiple times before but the proofs given haven't been given via the definition of uniform convergence but using sups and differentiation, and I've only come across the definition of uniform convergence so far.

Let $f_n:[0,1]→\mathbb{R}$ be a function defined by $f_n(x)=x^n(1−x)$

How would you prove that $f_n = x^n(1-x)$ converges to $0$ uniformly using the definition?

I've done the following so far:

Let $\epsilon>0$. Then $N >$_______ implies:


And I have no idea where to go from here. I've been given the hint that I may need a different argument for $x$ 'close' to 1 from other values of $x$ but I'm quite confused about this - is it trying to get at the fact that if $x$ is close to $1$, then $(1-x)$ is close to $0$?

Thank you in advance!

  • 1
    $\begingroup$ Your should write clearly. For example, let $f_n :[0,1]\rightarrow \mathbb{R}$ be a function defined by $f_n(x)=x^n(1-x)$. For your case, the domain of the function is crucial. $\endgroup$ – Danny Pak-Keung Chan Nov 9 at 14:27
  • $\begingroup$ The domain of $f_n$ can also be $\mathbb{R}$. However, in this case, the conclusion is completely different. $\endgroup$ – Danny Pak-Keung Chan Nov 9 at 14:28
  • $\begingroup$ Yes, the point is that $\lvert 1-x\rvert$ is small when $x$ is close to $1$. And for $x$ not so close to $1$, namely $0 \leqslant x \leqslant t(\epsilon)$, the factor $x^n$ can be made small by choosing $n$ large enough. $\endgroup$ – Daniel Fischer Nov 9 at 14:33
  • $\begingroup$ @DannyPak-KeungChan Sorry, I've just edited it! $\endgroup$ – Mathsical Studies Nov 9 at 14:45
  • $\begingroup$ Just show that $\sup(|f_n|)\to 0$. Rewriting it, you'll get the definition that $f_n$ converges to zero uniformly. $\endgroup$ – Jakobian Nov 9 at 14:54


The local maximum of the function happens at $x={n\over n+1}$ which is $$\max_{[0,1]} f(x)=\left({n\over n+1}\right)^n\cdot {1\over n+1}$$and use $$\lim_{n\to \infty}\left({n\over n+1}\right)^n={1\over e}$$


Given $1> \epsilon >0$, on the interval $(1-\epsilon, 1]$, we have $$ x^n(1-x) \leq 1^n (1-x) < \epsilon \ .$$

On the other hand there exists an $N$ such that for all $n>N$ $$ (1-\epsilon)^n < \epsilon ,$$ This is easy to see because $\epsilon$ is fixed and $(1-\epsilon)^n \to 0$ as $n \to \infty$. Now, for any $n>N$ and for any $x \in [0,1-\epsilon],$ $$ x^n(1-x) \leq (1-\epsilon)^n \cdot1 < \epsilon \ .$$

So, we succeeded in finding for every $\epsilon$, and $N$ such that if $n>N$, then $|f_n(x)-0| < \epsilon$ no matter where $x$ is in our domain. This is by definition equal to $f_n \to 0$ uniformly.


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