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I am trying to show that the sequence of functions $f_n(x) = x(1-x)^n$ converges uniformly to $0$ on $[0, 1]$.

Well at $0$ and $1$, $f_n(x) = 0$ for all $n$.

So let $x \in (0, 1)$.

$f_n(x)$ converges uniformly to the zero function if there exists a sequence of values $\alpha_n$ that converges to $0$ and $|f_n(x) - f(x)| < \alpha$ for all $x$ and all $n$.

Well let $1 - x < k < 1$.

Then we have that $(1-x)^n < (1-k)^n$ and $(1-k)^n \to 0$ as $n \to \infty$.

And as $x \in (0, 1) \implies x(1-x)^n < (1-k)^n$

Hence $f_n(x)$ converges uniformly to the zero function.

Is that correct?

sonicboom
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  • What about the $x$ with $1-x > k$? Also, have you heard of Dini's theorem? – Daniel Fischer Aug 09 '14 at 19:24
  • Check this. – Mhenni Benghorbal Aug 09 '14 at 19:32
  • @DanielFischer I don't know Dini's Theorem, we didn't cover that. Have I made some mistake in the above? I think you are saying that what I have come up with only works for a certain $x$, when it should work for all $x$, is that it? – sonicboom Aug 09 '14 at 19:53
  • Right, you only treat the $x$ with $1-x < k$ (and then you have $(1-x)^n < k^n$, not $< (1-k)^n$). Take a look at Dini's theorem, it should be in the book, otherwise wikipedia. It is quite useful. – Daniel Fischer Aug 09 '14 at 19:58
  • @DanielFischer Just looked at Dini's Theorem, we actually might have covered it in hindsight, although I don't think it was called that as such, as we didn't consider topological spaces or compact sets in this module. – sonicboom Aug 09 '14 at 20:52
  • @DanielFischer If I said $(1-x)^n < (1-k)^n$ would that then be valid or does it imply dependence on $x$? It seems that it implies dependence on $x$ and hence it doesn't satisfy the property required for uniform convergence? Or does it actually hold for all $x$, looking at it again, it seems that it does! – sonicboom Aug 09 '14 at 20:56

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Hint: Notice that:

$$f_{n}'(x)=(1-x)^{n}-nx(1-x)^{n-1}=(1-x)^{n-1}((1-x)-nx)=(1-x)^{n-1}(1-(n+1)x)$$

hence maximum at $x=\frac{1}{n+1}$ (Boundary terms evaluate to $0$).

user71352
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  • Thank you Daniel Fischer. I will correct my answer. – user71352 Aug 09 '14 at 19:29
  • Shouldn't we check the second derivative to determine if this is actually a maximum? – sonicboom Aug 09 '14 at 19:55
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    @sonicboom You could check that or you could notice that $1-x\ge0$ (with equality only at $x=1$) for $x\in[0,1]$ and that $(1-(n+1)x)>0$ for all $x\in[0,\frac{1}{n+1})$ and $(1-(n+1)x)<0$ for all $x\in(\frac{1}{n+1},1]$. So the function strictly increases from the left then strictly decreases to the right. – user71352 Aug 09 '14 at 19:59
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    @sonicboom Since the function is 0 at the endpoints and positive in between, it has to have its maximum in the interior. Since it's differentiable there, setting the derivative equal to 0 will detect the maximum. (The only possible difficulty would be if the derivative vanished at several places, and then you'd want to check which one is the maximum. But here the derivative vanishes only once in the interior of the interval.) – Andreas Blass Aug 09 '14 at 20:04