$L^p(\mathbb{R})$ contains a subspace which is is isometrically isomorphic to $\ell^p$

Question

For $1 \leq p \leq \infty$, is it true that $L^p(\mathbb{R})$ contains a subspace which is is isometrically isomorphic to $\ell^p$ ?

I don't think it's true. But I can't find any counterexamples.

Answer 1

Let $(E_n)$ be a partition of $\mathbb R$ such that $λ(E_n)<\infty$, where $λ$ denotes the Lebesgue measure. Consider the functions $ u_n = (\frac{1}{λ(E_n)})^{1/p} \ \textbf{1} _{E_n}$. Notice that $ ||{u_n}||_p =1 $ and $ {u_n} \mid_{E_j}=0$ for $j \neq n$. Show that for any $a_1, \dots , a_n \in \mathbb R$, $$ ||\sum_{k=1}^{n} a_k u_k||_p^p= \sum_{k=1}^n |{a_k}|^p .$$ Indeed, for $j>n$,
$$\left ({\sum_{k=1}^n a_k u_k } \right)\mid_{E_j}=0 $$ and for $1 \leq j \leq n$,

$$ ({\sum_{k=1}^n a_k u_k }) \mid_{E_j}=\frac{a_j}{\lambda(E_j)^{1/p}} .$$ Hence, $$||{\sum_{k=1}^{n} a_k u_k}||^p_{p} = \int |\sum_{k=1}^n a_k u_k|^p = \sum_{j=1}^{\infty} \int_{E_j} |{\sum_{k=1}^n a_k u_k}|^p = \sum_{j=1}^n \int_{E_j} \frac{|a_j|^p}{\lambda(E_j)} = \sum_{j=1}^n |{a_j}| . $$

Conclude that for any $ x =(x(n))_n \in \ell_p $, we have that $$ ||\sum_{n=1}^{\infty} {x(n) u_n}||^p_{p} = \sum_{n=1}^{\infty} |{x(n)}|^p $$ and thus the map $$ \ell_p \ni x \mapsto \sum_{n=1}^{\infty} x(n) u_n \in L_p $$ is an isometry.

Answer 2

Hint: consider functions that are constant on intervals...