Sequence of functions on $\mathcal{L}^1([0,1])$ with $\lim_{n\rightarrow \infty}||f_n||_1=0$ but $\sup\{f_n(x): n\in\mathbf{Z}^+\}=\infty$

lp-spacesmeasure-theoryreal-analysissequences-and-seriessupremum-and-infimum

Problem Statement.

Good evening. As the title suggests, I am having a hard time with the following exercise:

Prove that there exists a sequence $f_1,f_2,\ldots$ of functions on
$\mathcal{L}^1([0,1])$ such that $\lim_{n\rightarrow
\infty}||f_n||_1=0$ but $$\sup\{f_n(x): n\in\mathbf{Z}^+\}=\infty$$
for every $x\in [0,1].$

Notation.

Here $\mathcal{L}^1([0,1])$ means $\mathcal{L}^1(\lambda_{[0,1]})$ where $\lambda_{[0,1]}$ represents Lebesgue measure restricted to the Borel subsets of $\mathbf{R}$ that are contained in $[0,1]$.

Questions.

Ok, so my first question is how do I interpret $\lim_{n\rightarrow
\infty}||f_n||_1=0$? I know that by definition $$||f||_1=\int |f|\,d\mu$$ so I think this statement means $$\lim_{n\rightarrow \infty}||f_n||=\lim_{n\rightarrow\infty}\int|f_n|\,d\mu=0.$$ But I haven't been able to get anywhere with this, possibly because of my confusion with the other part of the question, which is even finding a sequence of functions that have the property $\sup\{f_n(x): n\in\mathbf{Z}^+\}=\infty$. I initially tried playing around with functions like $$f_n(x)=\frac{1}{\sqrt{nx}}$$ but this is undefined at $x=0$. So to summarize, my two questions are:

  1. Understanding the limit notation I pointed out.
  2. Finding a sequence of functions with an undefined supremum at every $x$.

Any suggestions towards either of these points of confusion would be wonderful.

Thank you for your time!

Best Answer

Here is an example of such a function: $$ f_n:=k\times 1_{[j/2^k,(j+1)/2^k]}, $$ where $n=2^k+j$ and $0\le j< 2^k$.

$$ f_1\equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \\ f_4=2\times 1_{[0,1/4]}, f_5=2\times 1_{[1/4,1/2]}, f_6=2\times 1_{[1/2,3/4]}, f_7=2\times 1_{[3/4,1]} \\ \ldots $$

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