I know that for a bounded $\Gamma$ it follows that $L^{q}(\Gamma) \subset L^{p}(\Gamma)$ if $q > p$. I have a few questions regarding how $L^{p}$ spaces relate with regard to convergence. Consider $L^{p}$ spaces $L^{p}(\Gamma)$ and $L^{p-\epsilon}(\Gamma)$ where $\Gamma$ is a bounded subset of $\mathbb{R}$ and $\epsilon \in (0,p-1]$. If $u_{n} \rightarrow u$ in $L^{p-\epsilon}(\Gamma)$ can it be shown that $u_{n} \rightharpoonup u$ weakly in $L^{p}(\Gamma)$?
Secondly, if we consider the same spaces as above, how could we show that if $u_{n} \rightharpoonup u$ in $L^{p}(\Gamma)$ then $u_{n} \rightharpoonup u$ in $L^{p-\epsilon}(\Gamma)$?
Thanks!
The answer to the first question is no. Let $\Gamma=[0,1]$ and $u_n(x)=x^{-1/p}\chi_{[n^{-2},n^{-1}](x)}$. Then for any $q\in(1,p)$ $$ \int_\Gamma|u_n(x)|^q\,dx=\int_{n^{-2}}^{n^{-1}}x^{-q/p}\,dx\le\frac{n^{1-q/p}}{1-q/p}\to0\text{ as }n\to\infty. $$ On the other hand $$ \int_\Gamma|u_n(x)|^p\,dx=\int_{n^{-2}}^{n^{-1}}x^{-1}\,dx=\log n\to\infty\text{ as }n\to\infty. $$ It follows that $u_n\to0$ in $L^q$ but $u_n\not\rightharpoonup0$ in $L^p$. Observe that $u_n$ and $u$ are in $L^p$.
The answer to the second question is yes, since $L^{q'}\subset L^{p'}$, that is, the dual of $L^q$ is contained in the dual of $L^p$. (As before $q\in(1,p)$.)