# Boundedness of a Hilbert space projection map

Reading this recent thread I was reminded of a related problem I still haven't solved so I post it here in hopes of a positive result.

Let $V_0 \subset H_0$ and $V_1 \subset H_1$ be separable Hilbert spaces with embeddings continuous, compact and dense. Let $$\phi_{1 \to 0}:H_1 \to H_0$$ be a bounded linear (bijective) map with bounded linear inverse $\phi_{0 \to 1}$. Suppose also that $$\phi_{1 \to 0}:V_1 \to V_0$$ is also bounded with the bounded inverse $\phi_{0 \to 1}$.

Let $\{b_j\}$ be an orthonormal basis for $H_0$ which is also orthogonal in $V_0$ and define the sets $$B^n_0 := \text{span}(b_1, ..., b_n)\quad \text{and}\quad B^n_1 = \phi_{0 \to 1}(B^n_0) = \text{span}(\phi_{0 \to 1}(b_1), ..., \phi_{0 \to 1}(b_n))$$

The orthogonal projection operator $P^n_0:H_0 \to B^n_0$ is defined to satisfy $(P^n_0u-u, u_n)_{H_0}=0$ for all $u_n \in B^n_0$, for $u \in H_0$. It's bounded in $H_0$ and let's assume that$$\lVert P^n_0u \rVert_{V_0} \leq \lVert u \rVert_{V_0}\quad\text{for all u \in V_0.}$$

Question: Define $P^n_1:H_1 \to B^n_1$ by $$(P^n_1v-v, v_n)_{H_1} = 0\quad\text{for all v_n \in B^n_1.}$$

Does this operator also satisfy $$\lVert P^n_1v \rVert_{V_1} \leq C\lVert v \rVert_{V_1}\quad\text{for all v \in V_1,}$$ for a constant $C$ not depending on $n$?

Remarks:

• Morally I feel like it should since there is a map between the two spaces which is continuous/bounded.

• Note that $\phi_{1 \to 0}(P^n_1v)=(\phi_{0 \to 1})^{-1}(P^n_1v) \in B^n_0$, so there is some $w \in V_0$ such that $P^n_0w = \phi_{1 \to 0}(P^n_1v)$, then by continuity of the maps we get $$\lVert P^n_1v \rVert_{V_1} \leq C\lVert w \rVert_{V_1}$$ but I can't get a bound on the $w$. Maybe it can be chosen in a way to get my estimate to work..

• It would be enough for me to consider the model case where the spaces $H$ are $L^2$-like spaces with $(u,v)_{H_0} := \int_\Omega uv$ and $(a,b)_{H_1} := \int_{\Omega} h\phi_{1 \to 0}a \phi_{1 \to 0}b$ where $h$ is a smooth weight (and $\Omega$ is bounded).