# Example of a sequence of functions

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.

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