# Problem of the Week #7 - July 16th, 2012

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#### Chris L T521

##### Well-known member
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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

##### Well-known member
Staff member
No one attempted the problem this week. Here's the solution (as provided by girdav).

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.

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