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I'm trying to prove the following.

Prove $f_n :[0,1] \rightarrow \Bbb R$ defined by $f_n (x)=x^n (1−x)$ converges uniformly to zero.

I know that for uniform continuity, we must find an $\varepsilon$ such that $|f_n(x)-0|<b_n<\varepsilon$. I'm having trouble proving $f_n(x) <1$. I've tried comparing it to $\frac{1}{n}$, $\frac{x}{n}$, and a couple more functions, but I can't seem to be able to actually show that $f_n(x)$ is less than those functions without making some wild assumptions.

EDIT: I should mention that I am not allowed to use the derivative, we have not proven that in class yet.

I'm needing a little direction with this proof. Any help would be appreciated, thank you.

NoVa
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  • What is the $\max$ of $f_n$? – copper.hat Mar 21 '17 at 04:39
  • I'm not allowed to take the derivative, we haven't done this in class yet. But, it cannot be greater than 1 due to our domain. – NoVa Mar 21 '17 at 04:40
  • The domain would preclude $x$ from being greater than 1, not $f_n(x)$. The maximum of $f_n$ is referring to its range, not its domain.

    To prove uniform convergence, you need to show that the maximum of $f_n$ tends to 0 as $n\to\infty$.

     – pre-kidney Mar 21 '17 at 04:42
  • @pre-kidney But how do you do this without using the derivative? – NoVa Mar 21 '17 at 04:44

2 Answers2

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Pick $\epsilon > 0$. Choose $\delta \in (0,1)$ such that if $\vert x - 1 \vert < \delta$, then $$\vert 1 - x \vert < \epsilon$$ and then choose $N \in \mathbb{N}$ such that $$(1 - \delta)^n < \epsilon$$ if $n > N$.

If $n > N$, then on $[1 - \delta, 1]$ we have $$\vert x^n(1 - x) \vert \leq \vert 1 - x \vert < \epsilon$$ and on $[0, 1 - \delta]$ we have $$\vert x^n(1 - x) \vert \leq \vert x^n \vert \leq (1 - \delta)^n < \epsilon.$$

Ethan Alwaise
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Here is an approach to bounding $x^n(1-x)$ for $0\leq x\leq 1$ without using derivatives. The AM-GM inequality says that the arithmetic mean of a sequence of positive numbers is greater than or equal to the geometric mean. Let us apply this to the $n+1$ numbers $x/n,\ldots,x/n$ and $1-x$. These numbers sum to 1, so the arithmetic mean is $1/(n+1)$. Computing the geometric mean gives $$ \frac{1}{n+1}\geq \sqrt[n+1]{(x/n)^n(1-x)}, $$ or in other words, $$ \frac{n^n}{(n+1)^{n+1}}\geq x^n(1-x). $$

This bound is uniform in $x$ and tends to 0 as $n\to\infty$, so it tells us that the functions $f_n=x^n(1-x)$ converge uniformly to 0 on the interval $[0,1]$.

pre-kidney
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  • Thank you, that was the inspiration I needed. I'll post my solution as soon as I finish it. – NoVa Mar 21 '17 at 05:01