# Inclusion of $L^p$ spaces

Let $X \subset L^1(\mathbb{R})$ a closed linear subspace satisfying \begin{align} X\subset \bigcup_{p>1} L^p(\mathbb{R})\end{align} Show that $X\subset L^{p_0}(\mathbb{R})$ for some $p_0>1.$

I guess the problem is that in infinite measure spaces the inclusion $L^p\subset L^q$ only holds for $p=q$. Is it maybe possbile to apply Baire's Theorem in some way?

-
To go further: does anyone have an example where $X$ is infinite-dimensional? –  D. Thomine Jul 7 '12 at 9:04
What do you think about defining $Y:=\{\lvert f \lvert\ \ \ \lvert \ \ f \in X\}\cup X$, then using your first proof to conclude that $Y\subset L^q$ and therefor also $X$. The only problem might be, to prove that $Y$ is also closed. –  Thomas Jul 10 '12 at 12:36

The comments posted below are related to a previous answer, which wasn't good. Now it's a new version, which is, I think, correct.

I think I have an approach which uses Baire's categories theorem. 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.

For an example where the space is infinite dimensional, look at the answers here.

A remark: we didn't use the fact that we worked on $\Bbb R$, and it seems it works for each measured space with a non-negative measure. That is, if $(S,\mathcal A,\mu)$ is a measured space with $\mu$ non-negative, and if $X$ is a closed subspace of $L^1(S,\mu)$ contained in $\bigcup_{p>1}L^p(X,\mu)$, then we can find $p_0$ such that $X\subset L^{p_0}(X,\mu)$.

-
I think your solution also has the problem that $\lvert f_k\lvert$ does not have to be in $X$. –  Thomas Jul 6 '12 at 16:45
I tried to solve the problem that way, but I think that what we get is way too weak. If $X = \mathbb{L}^1$, we may get $r_{k_0} = + \infty$ (or be very large), but the approximation can fail to be bounded in $\mathbb{L}^p$ norm for any $p>1$. So, somehow, we have to use the assumption on $X$, and I feel like we are back to square one... –  D. Thomine Jul 6 '12 at 21:19
@D.Thomine You are right. The problem seems much harder than it looks, except I'm missing something. –  Davide Giraudo Jul 6 '12 at 23:06
@DavideGiraudo: I think the above comments are very confusing after your edit (the new solution looks good to me). Maybe you could write an additional comment that you have rewritten this post now - to avoid confusion for anyone reading it in the future? –  Sam Jul 14 '12 at 15:50
@Davide: I worked at it for a while and I couldn't come up with a way to salvage my proof. I think your approach above (Baire Category) is the correct one. Since I was only leaving my post up to see if it could be turned into a correct solution, I have now deleted it in light of the solution above. I've checked your work carefully and I agree that this works. Nice job! –  J. Loreaux Jul 17 '12 at 23:53