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Consider $1\le p, and $u_n$ be a sequence in $L^p(R^d)\cap L^r(R^d)$ which is bounded in $L^r(R^d)$. If we consider that there exists $u \in L^p(R^d)$ such that $u_n \to u$ strongly in $L^p(R^d)$ as $n \to\infty$ . I would like to show that $u_n\in L^q(R^d)$ and this sequence converges strongly to $u$ in $L^q(R^d)$ as $n\to \infty$ . I want to solve this but i have no clue where to start :( . Thank you for your help . It looks like interpolation inequality can help, but still i am clueless.

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We can write $\alpha p+(1-\alpha)r=q$ for $\alpha\in (0,1)$, hence by Hölder's inequality, for $v\in L^p\cap L^r$ $\int |v|^q=\int |v|^{\alpha p}|v|^{(1-\alpha)r}\leq \left(\int |v|^p\right)^{1/\alpha}\left(\int |v|^r\right)^{1/(1-\alpha)}$ hence $\tag{1}\lVert v\rVert_{L^q}\leq \lVert v\rVert_p^{\frac p{q\alpha}}\cdot \lVert v\rVert_r^{\frac r{q(1-\alpha)}}.$ The fact that $u_n\in L^q$ follows from (1).

We have that $u\in L^r$: indeed, let $u_{n_k}$ an almost everywhere convergent subsequence (to $u$). We have by Fatou's lemma $\int |u|^r=\int\liminf_k|u_{n_k}|^r\leq \liminf_k\int|u_{n_k}|^r\leq \sup_n\lVert u_n\rVert_r^r<\infty.$

Still thanks to (1), we can write, because $u-u_n\in L^p\cap L^r$ that $\lVert u-u_n\rVert_q^q\leq \lVert u-u_n\rVert_p^{\frac p{\alpha}}\cdot \lVert u-u_n\rVert_r^{\frac r{1-\alpha}}\leq\lVert u-u_n\rVert_p^{\frac p{\alpha}}\left(\lVert u\rVert_r+\sup_k\lVert u_k\rVert_r\right)^{\frac r{1-\alpha}}.$ Since $\left(\lVert u\rVert_r+\sup_k\lVert u_k\rVert_r\right)^{\frac r{1-\alpha}}$ is a real constant, we are done.