(Motivation is below) Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$. Let $p\in [1,\infty)$ and $u\in L^p(\Omega)$. Is there any $q>p$ such that $u\in L^q(\Omega)$?
I already know that if this is true for some $p_0$, then this is true for every $p$. For suppose the above is valid for some $p_0$ and let $u\in L^p(\Omega)$. Then $u^{p/p_0}\in L^{p_0}(\Omega)$, and so $u^{p/p_0}\in L^q(\Omega)$ for some $q>p_0$. Hence $u\in L^{p\frac{q}{p_0}}(\Omega)$, and $p(q/p_0)>p$. Another fact that easily follows from $\Omega$ being bounded is that if $u\in L^p(\Omega)$ and $u\in L^q(\Omega)$, then $u\in L^r(\Omega)$ for every $p<r<q$.
I got to this question while studying the following: Let $\Omega$ be an open (not necessarily bounded) subset of $\mathbb{R}^n$, and let $\mathscr{D}'(\Omega)$ be the space of distributions in $\Omega$. We know that there is a canonical map $T:L^1_{loc}(\Omega)\rightarrow\mathscr{D}'(\Omega)$ given by $T(u)(\phi)=\int_\Omega u\phi dx$ for every $\phi\in\mathscr{D}(\Omega)= C_c(\Omega)$ ($\mathscr{D}(\Omega)$ is given a certain direct limit topology, as usual). It's easily shown that this map is well-defined and continuous relatively to the weak-* topology in $\mathscr{D}'(\Omega)$. The question is: is this mapping injective?
Suppose that the answer to the problem is yes, and let $u\in L^1_\text{loc}(\Omega)$ such that $T(u)=0$. Let $\widetilde{\Omega}\Subset\Omega$. Then $\widetilde{u}=u|_\widetilde{\Omega}\in L^1(\widetilde{\Omega})$and then $\widetilde{u}\in L^p(\widetilde{\Omega})$ for some $p>1$. Now we can see $\widetilde{u}$ as an element of the dual of some $L^{p'}(\widetilde{\Omega})$, $1<p'<\infty$. But since $\int_\widetilde{\Omega} u\phi dx=0$ for every $\phi\in C_c(\widetilde{\Omega})$, which is dense in $L^{p'}(\Omega)$, this means that $\widetilde{u}=0$
"The question is: is this mapping injective?" yes, by the fundamental lemma of the calculus of variations, and by the definition of $L^1$.