would really appreciate help in understanding pointwise convergence and uniform convergence.

For example for the questions below, how does f_n converge pointwise to f(x)=0? To see whether a function converges pointwise or not, don't you just take the limits of the sequence of functions f_n. Because wouldn't lim(f_n(x))=infinity rather than than 0?

Could someone please also explain how one would check for uniform convergence. In this example, the function would be uniformly continuous since the function is defined over a compact set (closed and bounded interval) and is pointwise convergent. But is there another way to check for uniform convergence.

Your help would be greatly appreciated!

Thank you!

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For pointwise convergence, you fix $x\in [0,1] $ and you compute $\lim_ {n\to+\infty}f(x) $. There will be pointwise convergence in the set containing $x $ for which the limit exists $(\in\mathbb R) $. in your example,

for $x=0$, the limit is zero. for $x=1$, it is zero. for $0 <x <1$, write $(1-x)^n=e^{n\ln (1-x)} $ and you will find zero since exponential is faster than polynomial.

thus all the limits are zero in $[0,1] $ . $(f_n) $ converges (pointwise) to function $0$ at $[0,1] $

For uniform convergence

find the maximum of $|f_n (x)-0|=f_n (x)$ at $[0,1] $.

$$f'_n (x)=n^2 (1-x)^{n-1}(1-x-nx) $$

it is attained at $$x_n=\frac {1}{n+1} $$ $$f (x_n)=n (\frac {n}{n+1})^{n+1}$$

$$\lim_{n\to+\infty}f_n (x_n)=+\infty$$

the convergence is not uniform at $[0,1] $.

  • $\begingroup$ Thank you very much for your help! Would it also be correct to look at the example in the following way: At x=0, limit=0 and at x=1, limit=0, therefore by Sandwich/Squeeze Theorem, limit for 0<x<1 is also 0? Therefore, the sequence of functions converges pointwise to the zero function. Thank you! $\endgroup$
    May 7 '17 at 11:19
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    $\begingroup$ @USERMATHS I edited the uniform convergence. hope it will help. $\endgroup$ May 7 '17 at 13:55
  • $\begingroup$ $$f_n(x_n)\sim\frac{n}e\to\infty$$ $\endgroup$
    – Did
    May 7 '17 at 14:26
  • $\begingroup$ @Did Yes thanks. i edited. $\endgroup$ May 7 '17 at 14:32
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    $\begingroup$ "The answer is that the sequence is not uniformly continuous." This is not the answer (by the way, what does this even mean to say that a sequence is or is not "uniformly continuous"?). $\endgroup$
    – Did
    May 7 '17 at 14:35

For $\:x_n=1/n\in[0,1],$ $$\lim_{n\to\infty}f_n\left(1/n\right)=\lim_{n\to\infty}n\cdot e^{-1}=\infty$$

Thus, $$\exists\varepsilon>0\:\:\forall N_{\varepsilon}\:\:\exists n\geqslant N_{\varepsilon}\:\:\exists x_n\in[0,1]:\left|f_n(x_n)-f(x_n)\right|>\varepsilon.$$

In this case, with such a $\:x_n,\:$ any fixed positive $\:\varepsilon\:$ will satisfy the proposition.


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