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Let $f_n(x) = x(1 - x)^n$. Find $f(x) = lim \: f_n(x)$ Determine or not the convergence is uniform on [0,1].

Okay so first of all we find that the point wise limit is zero and this function we expect intuitively that it will converge uniformly.

We know that $f_n$ will converge uniformly to f iff $\lim_{x \to \infty} [sup: {|f(x) - f_n(x)|\thinspace x \in S}] = 0 $

So we can just take the derivative and we find $x = 1$ or $x = 1/(n + 1)$ and we see indeed both will go to zero as n --> infinity, Hence it will converge uniformly.

I wanted to also solve this question using the regular definition of uniform convergence to get some practice, I started as follows $|x(1 - x)^n - 0| <= x^n$ and I am trying to find N that depends on n only and isolate x somehow but kinda of lost if someone could point me to right direction that would be amazing I just need more practice for uniform convergence using regular definition, i.e proving uniform convergence using regular definition.

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I can think of two nice approaches to this problem.

The first is very direct: you simply identify the maximum of $f_n$ with calculus, and notice that it goes to zero.

The second is more complicated, but can be applied to more problems. (Typically we cannot compute the maximum explicitly.) Here we are given $\varepsilon$ and want to choose $n$ to make $f_n$ uniformly below $\varepsilon$. The idea is that, when $n$ is large, $f_n$ is small due to the contribution of different terms in different parts of $[0,1]$. Away from $0$, everything is driven by the $(1-x)^n$ part; near $0$, everything is driven by the $x$ part. To make this precise, we'll break the interval up between $[0,\varepsilon]$ and $[\varepsilon,1]$. (In similar but more complicated problems we might break it up into $[0,\delta]$ and $[\delta,1]$, and then choose $\delta$ with $\varepsilon$.)

On the first interval everything is free: $x(1-x)^n \leq \varepsilon$ if $x \leq \varepsilon$, since $(1-x)^n \leq 1$.

On the second interval we have to work a little: there we have $x \leq 1$ so we need to choose $n$ in order to make $(1-x)^n < \varepsilon$ when $x \geq \varepsilon$. We can do this with logarithms. I'll leave the details to you.

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  • $\begingroup$ I see ye makes sense I like this idea of breaking the intervals up will use this when solving problems didn't see it before. $\endgroup$
    – user111750
    Dec 6 '14 at 6:22

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