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I am having a bit of trouble on this revision question.

To determine pointwise convergence: $\lim_{n\rightarrow\infty} = nx(1-x)^n $. For $x=0, x=1$, it's clear that the limit is $0$. How can I determine the limit for $0 \le x \le 1$? (My limit finding skills are rusty). I'm quite sure it is $0$ as well, but how do I go about showing this properly?

Supposing this is true, then to determine uniform convergence, let $d_n(x):= |f_n(x)-f(x)|= nx(1-x)^n - 0$

Then $$ d_n'(x)= n(1-x)^n - n^2(1-x)x = 0$$ if $x=1$. So the maximum of $f_n(x)$ occurs at $x=1$. It follows that:

$$0 \le d_n(x)= |f_n(x)-f(x)|< d_n(1) = 0$$ So $(f_n)$ converges uniformly on $[0,1]$. Would this be correct?

Thanks for the help in advance!

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    $\begingroup$ No, otherwise you could move limit inside the integral, but $\displaystyle\lim\limits_{n\to\infty}\int f_n=\lim\limits_{n\to\infty}\frac{n}{2n+2}=\frac{1}{2}\neq0=\int \lim\limits_{n\to\infty} f_n$ $\endgroup$ – OBDA May 27 '14 at 15:51
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The plan is good, the execution isn't.

If you maximize $f_n(x)$ for fixed $n$ as a function of $x$ you get $x=1/(n+1)$. Uniform convergence then doesn't hold because

$$\sup_x f_n(x) = \frac{n}{n+1} \Bigl( 1 - \frac{1}{n+1}\Bigr)^n \to \frac{1}{e},$$

as $n\to\infty$.

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    $\begingroup$ You're missing the $n$ in front of $f_n$ $\endgroup$ – Omnomnomnom May 27 '14 at 15:42
  • $\begingroup$ noted, fixed... $\endgroup$ – JPi May 27 '14 at 19:06
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You are correct in stating that the pointwise limit is zero.

Your differentiation, however, was incorrect. We have $$ d_n'(x)= n(1-x)^n - n^2(1-x)^{n-1}x = 0 \implies\\ n(1-x)^{n-1}((1-x) - nx) = 0 \implies\\ 1 - (n+1)x = 0 \implies\\ x = \frac 1{n+1} $$ So, the maximum difference is $$ n \cdot \frac{1}{n+1} \cdot \left( 1 - \frac{1}{n+1} \right)^n \not \to 0 $$ Which, in fact, shows that we do not have uniform convergence.

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  • $\begingroup$ Thanks, I realised I made an error in differentiation. $\endgroup$ – Terrence J May 27 '14 at 22:57
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To answer your first question about pointwise convergence for $0<x<1$, note that for a fixed value of $x$ there exists $y>0$ such that

$$(1-x) = \frac{1}{1+y}.$$

Hence,

$$0<nx(1-x)^n = \frac{nx}{(1+y)^n}.$$

Applying the binomial theorem we see that

$$(1+y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + \ ... \ >\frac{n(n-1)}{2}y^2,$$

and

$$0<nx(1-x)^n < \frac{2x}{(n-1)y^2}.$$

The RHS of the above inequality can be made smaller than any $\epsilon>0$ for fixed values of x and y by choosing a sufficiently large value of $n$.

In particular, let $N$ be an integer such that

$$N > 1+\frac{2x}{\epsilon y^2}.$$

Then for all $n \geq N$ we have

$$0<nx(1-x)^n < \epsilon,$$

and the pointwise limit of the sequence is $0$.

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  • $\begingroup$ Thanks, really helpful. Just a quick question, how did you derive your first line? $\endgroup$ – Terrence J May 27 '14 at 22:54
  • $\begingroup$ Is $y$ chosen, dependent on $x$ I suppose? Sorry, it just doesn't seem obvious to me. $\endgroup$ – Terrence J May 27 '14 at 23:06
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    $\begingroup$ This is a standard argument to show pointwise convergence. We consider $x$ as a fixed number between $0$ and $1$. It is true that $y$ depends on $x$, but it is independent of $n$ and must be greater than $0$ if $x$ is less than $1$. I'll show more detail in applying the binomial theorem to obtain the inequality. $\endgroup$ – RRL May 27 '14 at 23:29
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    $\begingroup$ After $$0<nx(1-x)^n < \frac{2x}{(n-1)y^2}.$$ Is the rest necessary? Could we not just say that since as $n \rightarrow \infty$ by squeeze it tends to $0$ also. So since for $x \in [0,1]$, $nx(1-x)^n$ tends to $0$, then we can conclude this is the pointwise limit? $\endgroup$ – Terrence J May 28 '14 at 2:28
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    $\begingroup$ @user153663: Yes you can apply the Squeeze Theorem to reach the same conclusion. $\endgroup$ – RRL May 28 '14 at 2:39
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Observe that $$\sup_{x \in [0,1]} |f_n(x)-f(x)| \geq \left|f_n \left(\frac{1}{n} \right) -f\left( \frac{1}{n} \right) \right|=\left(1-\frac{1}{n} \right)^n \to \frac{1}{e}$$

as $n \to \infty$ .So that we can't have $\lim_{n \to \infty} \sup_{x \in [0,1]} |f_n(x)-f(x)|=0$. i.e. there is no uniform convergence.

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There is a way, a bit redundant, because it uses the fact that $n^{1/n} \to 1$ when $n \to \infty$. Anyway, here is it.

For, $x=0$ or $x=1$, we don't have anything to do. Note that, for $0<x<1$, we have that $$f_n(x)^{1/n}=n^{1/n}x^{1/n}(1-x) \to 1-x,$$ when $n \to \infty$. So, for each $\epsilon>0$, there is $N \in \mathbb{N}$, such that $$ n>N \Rightarrow (1-x)-\epsilon<f_n(x)^{1/n}<(1-x)+\epsilon.$$ Take then $\epsilon>0$ such that $(1-x)+\epsilon <1$. It follows that $$ n>N \Rightarrow f_n(x)<[(1-x)+\epsilon]^n<1.$$ Passing $n \to \infty$, we have that $f_n(x) \to 0$.

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