Noncompactness of $L^{p}$ space in $[-1,1]$ interval I want to show that for $1\leq p < \infty$, $L^{p}([-1,1])$ is not compact. I have found an example here  ( Unit sphere in $L^p([0,1])$ is not compact. )by constructing the following sequence
\begin{equation}
u_{j}(x) := 
\begin{cases}
2^{\frac{k}{p}}, \quad \text{if }\frac{l}{2^{k}}\leq x \leq \frac{l+1}{2^{k}}\\
0, \quad \text{otherwise}
\end{cases}
\end{equation}
Here, $j = 2^{k} + l$ and $0\leq l < 2^{k}$ with $k= 0,1,2,...$.
It is trivial to see $\forall j \in \mathbb{N}, ||u_{j}||_{L^{p}} = 1$ and thus the sequence is uniformly bounded in $L^{p}$. Now, I want to claim that $\exists \varepsilon > 0\, \forall i,j \in\mathbb{N}, i\neq j, \,||u_{i}-u_{j}||_{L^{p}}\geq \varepsilon$
Without loss of generality, I assume $i < j$ and $i = 2^{k_{i}}+l_{i}$ and $j = 2^{k_{j}}+l_{j}$. For the special case $k_{i} = k_{j}$, I can easily obtain $||u_{i}-u_{j}||_{L^{p}}= 2^{\frac{1}{p}}\geq 1$
Next, I consider the case $k_{i}<k_{j}$. if $[\frac{l_{i}}{2^{k_{i}}},\frac{l_{i}+1}{2^{k_{i}}}]\cap[\frac{l_{j}}{2^{k_{j}}},\frac{l_{j}+1}{2^{k_{j}}}]=\emptyset$, then it is also not difficult to show that $||u_{i}-u_{j}||_{L^{p}}= 2^{\frac{1}{p}}\geq 1$
My problem is that I cannot show similar bound for $[\frac{l_{i}}{2^{k_{i}}},\frac{l_{i}+1}{2^{k_{i}}}]\cap[\frac{l_{j}}{2^{k_{j}}},\frac{l_{j}+1}{2^{k_{j}}}]\neq\emptyset$
For simplicity, I start from the case $\frac{l_{j}}{2^{k_{j}}}<\frac{l_{i}}{2^{k_{i}}}<\frac{l_{j}+1}{2^{k_{j}}}<\frac{l_{i}+1}{2^{k_{i}}}$
I am stuck at this part :
$$||u_{i}-u_{j}||_{L^{p}}^{p} = 2^{k_{j}}\big(\frac{l_{i}}{2^{k_{i}}} - \frac{l_{j}}{2^{k_{j}}} \big)+ \bigg|2^{\frac{k_{j}}{p}} - 2^{\frac{k_{i}}{p}} \bigg|^{p}\big(\frac{l_{j}+1}{2^{k_{j}}} - \frac{l_{i}}{2^{k_{i}}} \big)+2^{k_{i}}\big(\frac{l_{i}+1}{2^{k_{i}}} - \frac{l_{j}+1}{2^{k_{j}}} \big)$$
I dont know how to proceed from that line above to obtain the desired lower bound. Any help will be much appreciated! Thank you very much!
 A: Denote
$$
I_{k,l} = \left[\frac{l}{2^{k}},\frac{l+1}{2^k} \right],
$$
where $0\leq l<2^k$. The important observation is that if you have two of these intervals, $I_{k,l}$ and $I_{r,j}$, then either $I_{k,l}\cap I_{r,j} = \emptyset$ or $I_{k,l} \subset I_{r,j}$ if $k\geq r$. Your sequence is defined by $u_i = 2^{k/p}\chi_{I_{k,l}}$, where $i=2^{k_i}+l_i$. If $i\not= j$, assume $k_i\geq k_j$ without loss of generality. Then, we have two cases: if the intervals are disjoint, then
$$
\|u_i-u_j\|^p_p = \int_{I_{k_i,l_i}}2^{k_i}dx + \int_{I_{k_j,l_j}}2^{k_j}dx = 2.
$$
If they are not disjoint, then $I_{k_i,l_i} \subset I_{k_j,l_j}$ and we get
$$
\begin{align*}
\|u_i-u_j\|^{p}_p &= \underbrace{\int_{I_{k_i,l_i}}|2^{k_i/p}-2^{k_j/p}|^pdx}_{\geq 0} + \int_{I_{k_j,l_j}\setminus I_{k_i,l_i}}2^{k_j}dx \\
&\geq 1-\frac{1}{2^{(k_i-k_j)/p}}.
\end{align*} \\ 
$$
Note however, that $k_i>k_j$ in this case, because if $k_i=k_j$, the intervals $I_{k_i,l_i}$ and $I_{k_j,l_j}$ would not be disjoint (we are assuming $i\not=j)$. Therefore, $k_i-k_j \geq 1$, and 
$$
\|u_i-u_j\|^p_p \geq 1-\frac{1}{2^{1/p}}>0.
$$
It follows than that in any case, if $i\not= j$, $\|u_j-u_j\|_p \geq \min\{2^{1/p},(1-2^{-1/p})^{1/p}\}>0$.
