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 $$ p (u) \le K_1 d(u, N) \quad \forall u \in V . $$
- Prove that there is a constant $K_2 >0$ such that $$ K_2 d(u, N) \le p (u) \quad \forall u \in V. $$
There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it? My proof is by contradiction. Is there a direct proof?
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 $0 <c_1 < c_2 < c_3 < \infty$ such that $$ |u| \le c_1 \|u\| \le c_2 [u] \le c_3 \|u\| \quad \forall u \in V . \tag{$*$} $$
3.
Assume the contrary that for each $n \in \mathbb N^*$, there is $u_n \in V$ such that $\frac{1}{n} d(u_n, N) > p (u_n)$. Because $\dim N <\infty$, there is $v_n \in N$ such that $\|u_n-v_n\| = d(u_n, N)$. Then $\frac{1}{n} \|u_n-v_n\| > p (u_n)$. We have $p(u_n) = p(u_n-v_n)$. Then $\frac{1}{n} \|u_n-v_n\| > p (u_n-v_n)$. Let $x_n := \frac{u_n-v_n}{\|u_n-v_n\|}$. Then $\|x_n\|=d(x_n, N)=1$ and $\frac{1}{n} > p (x_n)$.
Clearly, $p(x_n) \to 0$. Then $(x_n)$ is a Cauchy sequence w.r.t. $p(\cdot)$. Because $i$ is compact, we assume WLOG that there is $x \in H$ such that $|x_n-x| \to 0$. Then $(x_n)$ is a Cauchy sequence w.r.t. $|\cdot|$. Then $(x_n)$ is a Cauchy sequence w.r.t. $[\cdot]$. Because $[\cdot]$ is equivalent to $\|\cdot\|$, we get $(x_n)$ is a Cauchy sequence w.r.t. $\|\cdot\|$. Because $V$ is complete w.r.t. $\|\cdot\|$, there is $y \in V$ such that $\|x_n-y\| \to 0$. By $(*)$, we get $|x_n-y| \to 0$ and thus $x=y$.
Then $\|x_n-x\| \to 0$ and thus $p(x_n-x) \to 0$. We have $p(x) \le p(x_n-x) + p(x_n) \to 0$. Then $p(x)=0$ and thus $x\in N$. Then $d(x_n, N) \le \|x_n-x\| \to 0$, which contradicts the fact that $d(x_n, N)=1$. This completes the proof.