$L^p$ subspaces of $L^1$

Question

Suppose I have a closed subspace $S$ of $L^1([0,1])$ such that each $f\in S$ lies in some $L^p([0,1])$ space for $p>1$. How can I prove that $S$ itself lies in some $L^p([0,1])$ for $p>1$? I would like to apply some sort of Banach-Steinhaus argument but can't see how to proceed.

Answer 1

Let's prove this by contradiction. Assume that there exists some $\{p_n\}\subset \mathbb{R}$ such that $p_n$ decreases strictly to $1$ and such that there is an $f_n\in S$ such that $p_n = \sup\{p > 1 : f_n\in L^p([0, 1])\}$. Without loss of generality, let $\|f_n\|_{L^1([0, 1])} = 1$ and $f_n(x)\geq 0$ for all $x\in [0, 1]$ and $n\in \mathbb{N}$. Note that as $[0, 1]$ is a finite measure space, $L^q([0, 1])\subset L^p([0, 1])$ for all $1\leq p < q$. Then, consider $f(x) := \sum_{n=1}^{\infty} 2^{-n}f_n(x)$ ($f(x)$ may be infinite at some points $x\in [0, 1]$). We have by Lebesgue's monotone convergence theorem that $$\|f\|_{L^1([0, 1])} = \lim_{N\to \infty} \left\|\sum_{n=1}^N 2^{-n}f_n\right\|_{L^1([0, 1])} = \lim_{N\to \infty} \sum_{n=1}^N 2^{-n} = 1$$ Furthermore, $$f(x)\geq \sum_{n=1}^N 2^{-n}f_n(x)\geq 2^{-N}f_N(x)$$ for all $N$, so given any $p > 1$, we let $N$ be such that $p_N < p$. We then have that $$\|f\|_{L^p([0, 1])}\geq 2^{-N}\|f_N\|_{L^p([0, 1])} = \infty$$ Therefore, $f\notin L^p([0, 1])$ for any $p > 1$, which contradicts our precondition on $S$. Thus, there is no sequence $\{p_n\}\subset \mathbb{R}$ such that $p_n$ decreases strictly to $1$ and such that there is an $f_n\in S$ such that $p_n = \sup\{p > 1 : f_n\in L^p([0, 1])\}$. This implies that there is some $p > 1$ such that for all $f\in S$, $f\in L^p([0, 1])$ (as otherwise, we could construct a sequence $\{p_n\}$ as above). Therefore, $S\subset L^p([0, 1])$.

Answer 2

Consider the subspace generated by $f_p(x)=\frac{1}{x^{2/p}}$ for $p=2,3,4,...$.

This space cannot be contained in a single $L^p$.