Functional analysis, bounded operators One question, if $X$ and $Y$ are two Banach spaces and  $A : X \rightarrow Y$ is a linear injective open operator, then $A$ has to be bounded?
 A: Since $A$ is open, it is surjective, hence it has a linear inverse $B\colon Y \rightarrow X$. For an open set $O\subset X$ the set $B^{-1}(O) = A(O)$ is open, thus $B$ is continuous and consequently bounded. As a consequence of the open mapping theorem, also $A=B^{-1}$ has to be bounded.
Why does openness imply surjectiveness? Take an open set $O\subset X$ containing $0$, then also $A(O)$ is open and contains $0$. Now for any $y\in Y$, there is a $0\neq \lambda \in \mathbb{R}$(or $\mathbb{C}$) with $\lambda y \in A(O)\subset \text{range}(A)$. The range is a subspace, hence also $y =\lambda^{-1}\cdot (\lambda y) \in \text{range}(A)$, thus $A$ is surjective. 
A: A proper vector subspace of $Y$ cannot have nonempty interior. And we have by assumption that $A(X)$ is a vector subspace of $Y$ which is open, so $A(X)=Y$.
Now the result follows by Open Mapping Theorem: The openness of $A$ together with bijection imply that $A^{-1}:Y\rightarrow X$, then $A^{-1}:Y\rightarrow X$ is open, flipping over again we get the boundedness of $A=(A^{-1})^{-1}$.
