Considering the Lebesgue measure on $\mathbb{R}$, find $f_n(x)$ such that $|f_n|_1\rightarrow 0$ but $\limsup f(x)=\infty$ for all $x$.

Question

This does trouble me somewhat, as I understand $|f_n|_1\rightarrow 0$ implies there is a sub-sequence that converges a.e., but my attempt at an example is as follows:

First, we think of the typewriter series of functions $t_n(x)= 1_{[\frac{n-2^k}{2^k},\frac{n-2^k+1}{2^k}]}$, where $2^k\leq n<2^k+1$, which does not converge pointwise on $[0,1]$, but $\limsup t_n(x)=1$, so we then consider $kt_n(x)$, which are then each of length $k\frac{1}{2^k}$, and $\lim \frac{k}{2^k}=0$, so this still converges to $0$ in $L_1$, but now $\limsup kt_n(x)=\infty$ for all $x\in[0,1]$.

Now, we want to populate the entire real line with this function, so my thought is to "take turns" with which interval gets this function, so $g_1(x)=kt_1(x)$, $g_2(x)=kt_2(x)$, $g_3(x)=kt_1(x-1)$, and so on. I suppose my greatest concern with this is how to precisely index this. Could we say let $g_n(x)=kt_n(x-m)$, where $m\leq n< m+1$?

Even still, that only covers the positive case, so I suppose we merely have to reflect this to the negatives and say $G_n(x)=g_n(|x|)$ is our sequence of functions. Now we can say this still converges in $L_1$ because our integrals are just of size $2*\frac{k}{2^k}$?

Is this valid? The indexing concerns me somewhat as to how valid it is, and I can't find a sub-sequence that converges a.e. Would we just take the portions stuck at $0$? That is, the sequence $f_n(x)=n1_{[0,\frac{1}{2^n}]}$?

Answer 1

Take the typewriter function $t_{n}(x)$, scale it by its index as you have done, and distribute copies of the function over the real line. In other words, let $\{a_k\}_{k \in \mathbb{Z}}$ be a positive sequence of integers such that $\sum_{k \in \mathbb{Z}} \frac{1}{a_k} = 1$ and consider the functions

$$c_n(x) = \sum_{k \in \mathbb{Z}} \frac{nt_n(x-k)}{a_k},$$

Then by monotone convergence, $$ \int |c_n(x)| \, \mathrm{d}x = \sum_{k \in \mathbb{Z}} \int \frac{nt_n(x-k)}{a_k} \, \mathrm{d}x = \frac{n}{2^n}\cdot\sum_{k \in \mathbb{Z}} \frac{1}{a_k} = \frac{n}{2^n} \to 0.$$

Finally, given $x \in \mathbb{R}$, let $k = \lfloor x \rfloor$ and observe that $$\limsup c_n(x) \geq \frac{1}{a_k} \limsup nt_n(x - k) = \infty,$$ since $\limsup t_n(x-k) = 1$.

Answer 2

$g_{n,k} =\sqrt{n}$ on $[\frac{k}{n},\frac{k+1}{n}]$, $0$ outside, $\int g_{n,k} = \frac{1}{\sqrt{n}}$.

Arrange the $(g_{n,k})_{1\le n, -n\le k\le n-1}$ in any order and get a desired sequence.