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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?

user284331
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Negreanu
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2 Answers2

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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.

Jan Bohr
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  • Could you please help with the proof ? –  Jan 20 '18 at 22:58
  • What do you mean? – Jan Bohr Jan 21 '18 at 01:50
  • @jabo Thank you! And can we say that if we have X and Y two Banach spaces and A:X→Y linear injective CLOSED operator, then A is unbounded (there exist a constant c such that norm(Ax)>=c*norm(x))? – Negreanu Jan 21 '18 at 13:13
  • What do you mean by closed? If $A$ has a closed graph, then it is bounded by the closed graph theorem. If $A$ maps closed sets to closed sets in particular it has closed range. This is equivalent to the estimate $\Vert Ax\Vert \ge c \Vert x\Vert$ (also a consequence of the open mapping theorem). (And the estimate does not have anything to do with unboundedness.) – Jan Bohr Jan 21 '18 at 17:59
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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 $A=(A^{-1})^{-1}$.

user284331
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  • Hi! Thank you for your response. I have one question, using what you have said I have that A is bijective, linear and open. The open mapping theorem says that if an operator is surjective, continuous and linear then it is open, but if we have that is open surjective and linear we can say that is continuous? – Negreanu Jan 21 '18 at 12:13
  • The trick is that, the openness of $A$ together with bijection imply that $A^{-1}:Y\rightarrow X$ is bounded, then it is time to use Open Mapping Theorem to claim that $A^{-1}:Y\rightarrow X$ is open, flipping over again we get the boundedness of $A=(A^{-1})^{-1}$. – user284331 Jan 21 '18 at 18:57