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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?

Thomas
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    To go further: does anyone have an example where $X$ is infinite-dimensional? – D. Thomine Jul 07 '12 at 09:04
  • 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
  • This is exercise 4.8 on Brezis's book Functional Analysis, Sobolev Spaces and Partial Differential Equations – Evangelopoulos Foivos Jan 13 '23 at 15:23

1 Answers1

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I think I have an approach which uses Baire's category 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 category 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 $X\subset L^{p_0}(X,\mu)$.

Xander Henderson
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Davide Giraudo
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  • @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