0
$\begingroup$

Construct an example of a sequence of functions $(f_n)$ defined on $[0,1]$ such that $f_n$ converges pointwise to $0$ and for every sequence of numbers $(a_n)$ that tends to $\infty$, sequence $(a_nf_n)$ doesn't converge to $0$. Does anyone have any ideas? I am having trouble overcoming the fact we do not know the speed of $(a_n)$. What ever I come up with for $f_n$ I think I could find fast/slow enough $a_n$ to make $(a_nf_n)$ go to $0$. Obviously, this is not true. So, I would be very grateful for some help.

I found out this is an exercise in Rudin's Functional analysis. Chapter 1, exercise 7.

$\endgroup$
1
$\begingroup$

So, I made a mistake. The functions $f_n$ need not be continuous, just plane functions defined on $[0,1]$. Anyway, the collection of all complex sequences converging to 0 has the cardinality $c$, that is the same cardinality as $[0,1]$. We define $f_n(x)$ to be one of the sequences converging to $0$. We can do this for every $x\in[0,1]$, because of the fore mentioned bijection. This sequence of functions $(f_n)$ is pointwise convergent to $0$. But if we take any sequence $a_n\to\infty$, then $(a_nf_n(x_0))$ will not converge to $0$ for some $x_0\in[0,1]$. Among all those sequences there is also, for example, the sequence $\dfrac{1}{a_n}\to0$ and that will ensure we don't have pointwise convergence to $0$.

$\endgroup$
0
$\begingroup$

Maybe you can start with the sequence : $$f_n(x)=x^n; x<1$$ and $$f_n(1)=0$$ , and, like you suggested, multiply $f_n$ by some sequence that grows "fast-enough". You may want to consider that $d/dx(x^n)=nx^{n-1}<n$ to construct the sequence $a_n$

$\endgroup$
  • $\begingroup$ But your $f_n(x)$ are not continuous. And this has to be true for every $a_n\to\infty$. $\endgroup$ – Poppy Nov 13 '13 at 3:00
  • $\begingroup$ Actually both $x^n$ and $0$ are continuous everywhere, but I did not consider the second point. $\endgroup$ – user99680 Nov 13 '13 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.