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Could someone suggest an $n$-dependent but $x$-independent upper bound for $x^n(1-x^n)$ (used to show uniform convergence) on $[0,1]$? Thanks.

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Substitute $x^n = y$ so that you get the expression $y(1-y)$. Can you maximize this expression in $y$? –  Srivatsan Sep 15 '11 at 15:38
    
Ah, thanks!! I get it now :-) –  analysisnewbie Sep 15 '11 at 15:46
    
This, or the fact that $4x^n(1-x^n)=1-(1-2x^n)^2$. –  Did Sep 15 '11 at 16:34
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up vote 3 down vote accepted

Substituting $x^n =: y$, we get the function $g(y) = y(1-y)$. As $x$ varies over $[0,1]$, so does $y$. (I am assuming $n$ is a positive integer.)

So the maximum of that expression is the same as the maximum of $g(y)$ over $[0,1]$. By differentiation or AM-GM, you can show that $g(y)$ is maximized when $y = \frac{1}{2}$, and the maximum value is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. You can also find the $x$ that maximizes the given expression by substituting back $x^n$ for $y$ in $y = \frac{1}{2}$.

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