Compact injections and equivalent seminorms Let $V$ and $H$ be two Banach spaces with norm $\lVert \cdot \rVert$ and $\lvert \cdot \rvert$ respectively such that $V$ embeds compactly into $H$.  Let $p$ be a seminorm on $V$ such that $p(u) + \lvert u \rvert$ is a norm on $V$ that is equivalent to $\lVert \cdot \rVert$.  Set $N = \{u \in V: p(u) = 0\}$.  Prove that there does not exist a sequence $(u_n)$ in $V$ satisfying


*

*$\operatorname{dist}(u_n, N) = 1$ for all $n$

*$p(u_n) \to 0$.  


Ideas: I have no reason why I should expect such a result, so I can't motivate it.  Anyway, I want to claim that $u_n$ approach a limit $u$.  Then hopefully $p(u_n) \rightarrow p(u) = 0$ so $u \in N$, contradicting $1 = \operatorname{dist}(u_n,N) \rightarrow \operatorname{dist}(u,N) = 0$.  It would help greatly if the $(u_n)$ were bounded, since then the compact injection means that $u_n$ approach a limit in $\bar{V} \subset H$.  Then somehow argue that the limit is actually in $V$?

We will first show that, given $x\in X$, there exists $z\in N$ s.t. $\|x-z\|_X=d(x-z,N)$ and $p(x)=p(x-z)$. Let $\{z_n\}\subset N$ s.t. $\|x-z_n\|_X\to d(x,N)$. Then $\infty>\|x\|_X+\sup_{n}\|x-z_n\|_X>\sup_{m}\|z_m\|_X$. Since $N$ is finite dimensional, there exists a subsequence $z_{n_k}$ and $z\in N$ s.t. $z_{n_k}\to z$. Therefore, $\|x-z_{n_k}\|_X\to\|x-z\|_X=d(x,N)$. Moreover, $p(x-z)=p(x)$ because
\begin{align*}
 p(x)&\leq p(x-z)+p(z)=p(x-z)\\
 p(x-z)&\leq p(x)+p(-z)=p(x)
\end{align*}
and $d(x-z,N)=d(x,N)$ since $N$ is a subspace.
We now prove the main claim. AFSOC that for all $n\in\mathbb{N}$ there exists $x_n\in X$ s.t. $d(x_n,N)>np(x_n)\geq 0$. Normalizing $x_n$ by $d(x_n,N)$ we have that $1>np(x_n)$. For each $x_n$, let $z_n\in N$ be the $z$ found above. Then we have that $\|x_n-z_n\|=1$ for all $n\in\mathbb{N}$ so $\{x_n-z_n\}$ is a bounded sequence. By compactness of the embedding, $\{x_n-z_n\}$ has a subsequence, $\{x_{n_k}-z_{n_k}\}$ that converges to some $\hat{z}$. By continuity of $p$, we have that $p(x_{n_k}-z_{n_k})\to p(\hat{z})=0$. This implies that $\hat{z}\in N$. However, this implies that $d(x_{n_k}-z_{n_k},N)\leq\|(x_{n_k}-z_{n_k})-\hat{z}\|\to0$: a contradiction. 
