I was wondering whether a closable operator $A$ with domain $D(A)\subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
I was wondering whether a closable operator $A$ with domain $D(A)\subset H$ ($H$ is a Hilbert space) which is also bijective has an everywhere defined bounded inverse?
Thanks in advance.
Math.
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