The problems is this:
Let $X \subseteq Z$ and $Y \subseteq V$ with $X,Y,V,Z$ all Banach spaces, and the inclusions continuous. Let $P: Z \to V$ be a continuous linear map with the property that $$u \in Z, Pu \in Y \implies u \in X $$
\We want to show that there is a constant $C > 0$ such that for all $u \in X$ with $Pu \in Y$ we have:
$$ || u ||_X \leq C ( || Pu||_Y + ||u||_Z)$$
My current strategy is this:
Try to show that $X \cap P^{-1}(Y)$ is closed in $X$
Try to show that the map $x \to (x, Px)$ is continuous from $X \cap P^{-1}(Y)$ to $Z \oplus Y$ and that the image is closed.
Employ the open mapping theorem
But, I can't see any reason why $X \cap P^{-1}(Y)$ would be closed in $X$, nor why the image of the map above would be closed. If I assume that $X, Y$ are closed subspaces of $Z,V$ individually, then I can do it. But, I don't see any reason why they should be closed in general.
Please let me know if there is another stragegy or solution.
Let $G = \{(u, Pu) \in Z \oplus Y : u \in Z, Pu \in Y\}$. Verify that this is a closed subspace of $Z \oplus Y$, hence a Banach space in the norm $\|u\|_Z + \|Pu\|_Y$.
To see this, suppose $(u_n, P u_n) \in G$ converges in $Z \oplus Y$ to some $(u, v)$; this means $u_n \to u$ in $Z$ and $P u_n \to v$ in $Y$. A fortiori, we also have $P u_n \to v$ in $V$. Now we have to show that $(u,v) \in G$ which is to say that $v = Pu$. But the continuity of $P$ implies that $P u_n \to Pu$ in $V$, so by uniqueness of limits in $V$, we have $v = Pu$.
Now consider the linear operator $T : G \to X$ defined by $T((u,Pu)) = u$. Use the closed graph theorem to show that $T$ is continuous, and then you are done.