Can a Closed Subspace of $L^1(\Bbb R)$ be Contained in All $L^p(\Bbb R)$ Spaces?

[Chris L T521]

Many thanks to girdav for this week's problem!

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Problem: Let $X$ a closed subspace of $L^1(\Bbb R)$. We assume that $X\subset \bigcup_{1<p<\infty}L^p(\Bbb R)$. Show that we can find $p_0\in (1,+\infty)$ such that $X\subset L^{p_0}(\Bbb R)$.
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[Chris L T521]

No one attempted the problem this week. Here's the solution (as provided by girdav).

Spoiler

We define for an integer $k$$$F_k:=\{f\in X: \lVert f\rVert_{L^{1+1/k}}\leq k\}.$$- $F_k$ is closed (for the $L^1$ norm). Indeed, let $\{f_j\}\subset F_k$ which converges in $L^1$ to $f$. A subsequence $\{f_{j'}\}$ converges to $f$ almost everywhere, hence
$$\int_{\Bbb R}|f|^{1+1/k}dx=\int_{\Bbb R}\liminf_{j'}|f_{j'}|^{1+1/k}dx\leq
\liminf_{j'}\int_{\Bbb R}|f_{j'}|^{1+1/k}dx\leq k.$$
- We have $X=\bigcup_{k\geq 1}F_k$. Indeed, take $f\in X$; then $f\in L^p$ for some $p>1$. For $k$ large enough, $1+1/k\leq p$ and breaking the integral on the sets $\{|f|<1\}$, $\{|f|\geq 1\}$
$$\lVert f\rVert_{L^{1+1/k}}^{1+1/k}\leq \lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p,$$
so
$$\lVert f\rVert_{L^{1+1/k}}\leq \left(\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p\right)^{1-\frac 1{k+1}}.$$
The RHS converges to $\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p$, so it's smaller than two times this quantity for $k$ large enough. Now, just consider $k$ such that
$$2\left(\lVert f\rVert_{L^1}+\lVert f\rVert_{L^p}^p\right)\leq k.$$By Baire's categories theorem, we get that a $F_{k_0}$ has a non-empty interior. That is, we can find $f_0\in F_{k_0}$ and $r_0>0$ such that if $\lVert f-f_0\rVert_{L^1}\leq r_0$ then $f\in F_{k_0}$. Consider $f\neq 0$ an element of $X$. Then $f_0+\frac{r_0f}{2\lVert f\rVert_{L^1}}\in F_{k_0}$. We have that
$$\left\lVert \frac{r_0f}{2\lVert f\rVert_{L^1}}\right\rVert_{L^{1+1/k_0}}\leq
\left\lVert f_0+ \frac{r_0f}{2\lVert f\rVert_{L^1}}\right\rVert_{L^{1+1/k_0}}+\lVert f_0\rVert_{L^{1+1/k_0}}\leq 2k_0,$$
hence
$$\lVert f\rVert_{1+1/k_0}\leq \frac{4k_0}{r_0}\lVert f\rVert_{L^1},$$
which proves the embedding.