Let
- $U_i,H$ be $\mathbb R$-Hilbert spaces
- $A_i\in\mathfrak L(U_i,H)$ with $A_1U_1\subseteq A_2U_2$
Assume $A_1$ is injective. How can we show that $A_2^{-1}A_1$ is bounded?
Clearly, if $A_2U_2$ is closed, then $A_21^{-1}$ is bounded by the bounded inverse theorem and hence $A_2^{-1}A_1$ is bounded as well.
If $A_1U_1$ is not known to be closed, then we should be able to show the desired claim by the closed graph theorem. Therefore, let $(u_1^n)_{n\in\mathbb N}\subseteq U_1$ and $u_1\in U_1$ with $$\left\|u_1^n-u_1\right\|_{U_1}\xrightarrow{n\to\infty}0\tag1$$ and $u_2\in U_2$ with $$\left\|A_2^{-1}A_1u_1^n-u_2\right\|_{U_2}\xrightarrow{n\to\infty}0\tag2.$$
We need to show that $A_2^{-1}A_1u_1=u_2$. How can we do that?
We know that there is a sequence $(u_2^n)_{n\in\mathbb N}\subseteq U_2$ with $$A_1u_1=A_2u_2$$ and hence $$A_2^{-1}A_1u_1^n=u_2^n$$ for all $n\in\mathbb N$. However, that doesn't help, unless we would be able to choose $(u_2^n)_{n\in\mathbb N}$ such that $\left\|u_2^n-u_2\right\|_{U_2}\xrightarrow{n\to\infty}0$ ...
By applying $A_1$ to (1), you get $$A_1 u_1^n \to A_1 u_1.$$ By applying $A_2$ to (2), you get $$A_1 u_1^n \to A_2 u_2.$$ Hence, $$A_1 u_1 = A_2 u_2.$$