In functional analysis I have come across unbounded operators defined in the following way: $$A:\mathcal D(A)\subseteq X\to Y,$$ where $X,Y$ are Hilbert spaces and $A$ is defined only for $x\in\mathcal{D}(A)$. My question is does there exist an unbounded which is defined on the whole of $X$? If yes, can we construct such an example?
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1You may be interested in this question – MPW Mar 26 '19 at 08:05
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IIRC you need some variant of the axiom of choice to define such an operator $A$. You cannot write down any explicit example. – gerw Mar 26 '19 at 10:43
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On every infinite dimensional normed linear space there is a linear functional which is not continuous. Such a functional is defined everywhere but it is unbounded.
Hints for construction of an unbounded operator: There exists a Hamel basis $A$ consisting of unit vectors and $A$ is an infinite set. Let $(x_n)$ be a sequence in $A$. Let $f(x_n)=n$ and $f(x)=0$ if $x \in A$ with $x \neq x_n$ for all $n$. We can now extend $f$ to the whole space by liearity to gegt an unbounded operator.
Kavi Rama Murthy
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In the case where $A$ is a closed operator (which covers a large class in applications), if $\mathcal{D}(A)=X$ then by the closed graph theorem $A$ will be bounded.
Then you can try with non closed operators.
S. Maths
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