# Does Uniform Boundedness in the Sobolev Space $W^{1,2}$ and Convergence in $L^p$ $(1 \leq p < 6)$ Imply Convergence in $L^6$?

Let $B$ denote the open unit ball in $\mathbb{R}^3$. I want to either prove or disprove that a sequence of functions $u_m$ in the Sobolev space $W^{1,2}(B)$ which is uniformly bounded in the $W^{1,2}(B)$ norm and which is convergent in $L^p(B)$ for all $1 \leq p < 6$ must be convergent in $L^6(B)$.

Can someone please point me in the right direction?

I know (cf. Chapter 5 of Evans PDE book) that the bound $$\| u \|_{L^q(U)} \leq C(k,p,n,U) \| u \|_{W^{k,p}(U)}$$ holds when $U$ is a subset of $\mathbb{R}^n$ having smooth boundary, $u \in W^{k,p}$, $k < \frac{n}{p}$, and $\frac{1}{q} = \frac{1}{p} - \frac{k}{n}$. The constant $C = C(k,p,n,U)$ is independent of $u$.

It follows from this bound that the $u_m$ are uniformly bounded in $L^6(B)$.

Since $W^{1,2}(B)$ is a Hilbert Space, uniform boundedness of $u_m$ in $W^{1,2}(B)$ also implies existence of a subsequence $u_{m_j}$ that converges weakly in $W^{1,2}(B)$, hence weakly in $L^6(B)$.

Thanks!

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As bonnnnn2010 already pointed out, $6$ is the critical exponent where you don't have compact embedding of $H^1(B)=W^{1,2}(B)$ anymore, for dimension $d=3$. Here's an example of a bounded sequence in $H^1(B)$ that is convergent in $L^p(B)$ for $p<6$, but which has no convergent subsequence in $L^6(B)$: Take any $\varphi\ne0$ in $H^1(\mathbb{R}^3)$. Then the sequence $(\varphi_n)$ with $$\varphi_n(x) = n^{1/2}\varphi(nx)$$ has the desired properties; in fact, $\varphi_n\to0$ in $L^p(B)$ for $p<6$.
Because $L^6$ is critical to the compact embedding of $H^1$ when $d=3$, so we can not expect the convergence in $L^6$, I'm not sure if there's counter-example in the book of Adams "Sobolev Spaces".