Show that for every $p\geq 1,$ $\ell_p$ is linearly isometric to a subspace of $L_p[0,1]$.

Question

The following question is taken from the 'Banach Space Theory: The Basis for Linear and Nonlinear Theory,' page $34,$ question $1.18.$

Question: Show that for every $p\geq 1,$ $\ell_p$ is linearly isometric to a subspace of $L_p[0,1].$

A hint given in the book is as follows:

Consider span$\{f_n\},$ where $f_n(x) = (n(n+1))^{\frac{1}{p}}\chi_{[\frac{1}{n+1},\frac{1}{n}]}.$

For each natural number $n,$ define $$e_n = (0,...,0,\overset{n}{1},0,...).$$ Clearly $e_n\in\ell_p.$

My attempt: Based on the hint, I define a map $T:\{e_n:n\in\mathbb{N}\}\to \text{span}\{f_n:n\in\mathbb{N}\}$ by $$T(e_n) = f_n.$$ Since $\{f_n:n\in\mathbb{N}\}$ is linearly independent, so $T$ is well-defined. Observe that $T$ is an isometry. Indeed, for each natural number $n,$ $$\|f\|_{L_p} = \left( \int_{\frac{1}{n+1}}^{\frac{1}{n}} (n(n+1))\,dt \right)^{\frac{1}{p}} = 1 = \|e_n\|_{\ell_p}.$$ However, I do not know how to expand $T$ to both $\ell_p$ and $L_p[0,1].$ Any hint would be appreciated.

Answer 1

The key feature here is that the $f_n$ have disjoint support (and all have norm $1$), that makes the linear extension of the map sending $e_n$ to $f_n$ an isometry. For $1 \leqslant p < \infty$, the family $\{e_n : n \in \mathbb{N}\}$ is a Schauder basis of $\ell^p(\mathbb{N})$, thus we can write our map as

$$T\biggl(\sum_{n \in \mathbb{N}} c_n e_n\biggr) = \sum_{n \in \mathbb{N}} c_n f_n\,. \tag{1}$$

One of course still needs to verify that this is well-defined and an isometry, but that is straighforward.

For $p = \infty$ we cannot argue in exactly the same way since there the $e_n$ do not constitute a Schauder basis, and for $y \in \ell^{\infty}(\mathbb{N})$ in general the series

$$\sum_{n \in \mathbb{N}} y_n e_n \qquad \text{and} \qquad \sum_{n \in \mathbb{N}} y_n f_n$$

aren't convergent in $\ell^{\infty}(\mathbb{N})$ and $L^{\infty}([0,1])$ respectively. These series are convergent [with respect to the norm topology] if and only if $y$ belongs to the subspace $c_0(\mathbb{N})$ of sequences converging to $0$.

We can still use $(1)$ in the case $p = \infty$ if we also consider the weak$^{\ast}$ topology on $\ell^{\infty}/L^{\infty}$ as the dual of $\ell^1/L^1$, for the series converge in the weak$^{\ast}$ topology, and every $y \in \ell^{\infty}$ can be written as $\sum c_n e_n$ in a unique way. It is however rather ugly to drag the weak$^{\ast}$ topology into the discussion of an isometry. But the alternative isn't too pretty either. We can specify the isometry $T \colon \ell^{\infty}(\mathbb{N}) \hookrightarrow L^{\infty}([0,1])$ explicitly as

$$T(y)\colon x \mapsto \begin{cases} 0 &, x = 0 \\ y_n &, \frac{1}{n+1} < x \leqslant \frac{1}{n} \end{cases}\,. \tag{2}$$

One verifies that this indeed is an isometry. We can also specify $T$ in the style of $(2)$ for $p < \infty$, with $T(y)(x) = y_n\cdot \bigl(n(n+1)\bigr)^{1/p}$ for $\frac{1}{n+1} < x \leqslant \frac{1}{n}$ if we want a uniform treatment of the cases $p = \infty$ and $p < \infty$.

Answer 2

Here is the same proof written without Schauder base of $l_p$. For the case $1 \leq p < \infty$ let $T : l_p \rightarrow L_p$ be the operator defined by $$Tx = \sum_{n=1}^\infty \frac{x_n}{\mu(A_n)}1_{A_n}$$

where $x=(x_n)_{n\in \mathbb N^*}$ , $A_n = ]\frac{1}{n+1}, \frac{1}{n}]$ for all $n \in \mathbb N^*$ and $\mu$ the Lebesgue mesure associated to $L_p$.

Let's note that the sets $A_n$ are disjoint and that $[0,1]=\cup_{n=1}^\infty A_n$

$T$ is a linear operator due to the linearity of the sum operator. Let's prove that $T$ is an isometry.

$$\|Tx\|_{L_p}^p = \int_{\cup_{n=1}^\infty A_n} |\sum_{m=1}^\infty \frac{x_m}{\mu(A_m)}1_{A_m}|^p d\mu = \sum_{n=1}^\infty \int_{A_n} |\sum_{m=1}^\infty \frac{x_m}{\mu(A_m)}1_{A_m}|^p d\mu$$ and $$\|Tx\|_{L_p}^p= \sum_{n=1}^\infty \int_{A_n} |\frac{x_n}{\mu(A_n)}1_{A_n}|^p d\mu = \sum_{n=1}^\infty |x_n|^p = \|x\|_{l_p}^p$$

For the case $p = \infty$, the proof is similar with $$Tx = \sum_{n=1}^\infty x_n 1_{A_n}$$