I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $(V, \| \cdot \|)$ and $(H, | \cdot |)$ be real Banach spaces satisfying $V \subset H$ with compact injection. Let $p(\cdot)$ be a seminorm on $V$ such that $p(\cdot) + |\cdot|$ is a norm on $V$ that is equivalent to $\| \cdot \|$. Let $N := \{u \in V : p(u)=0\}$ and $$ d(u, N) := \inf _{v \in N}\|u-v\| \quad \forall u \in V . $$
- Prove that $N$ is finite-dimensional.
- Prove that there is a constant $K_1 >0$ such that $$ p (u) \le K_1 d(u, N) \quad \forall u \in V . $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? I'm also happy to see other approaches.
We have $i:(V, \|\cdot\|) \to (H, |\cdot|), u \mapsto u$ is a compact (bounded linear) operator. Let $[\cdot] := p(\cdot) + |\cdot|$. The hypotheses imply that $[\cdot]$ is a norm on $V$ and that there are $c_1, c_2, c_3>0$ such that $$ |u| \le c_1 \|u\| \le c_2 [u] \le c_3 \|u\| \quad \forall u \in V . \tag{$*$} $$
1.
Clearly, $N$ is a vector subspace of $V$ and thus of $H$. Let $B$ be the closed unit ball of $N$ w.r.t. $|\cdot|$. It suffices to prove that $B$ is compact w.r.t. $|\cdot|$. We have $[u] =|u|$ for $u \in B$. By $(*)$, $B$ is bounded w.r.t. $\| \cdot \|$. Because $i$ is compact, $B$ has compact closure w.r.t. $|\cdot|$. The claim then follows.
2.
From $(*)$, we have $$ p(u-v) \le [u-v] \le \frac{c_3}{c_2} \|u-v\| \quad \forall u,v \in V . $$
Then $$ p(u) \le \frac{c_3}{c_2} \|u-v\| \quad \forall u \in V, v\in N . $$
Then $$ p(u) \le \frac{c_3}{c_2} \inf_{v\in N} \|u-v\| \quad \forall u \in V. $$
Then $$ p(u) \le \frac{c_3}{c_2} d(u, N) \quad \forall u \in V. $$