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I am self-studying the chapter of closed and closable operators. I have the following problem which I cannot find its proof.

Let $A$ be a closable operator and denote by $B$ a closed extension. We call $\bar{A}$ the closure of the (closable) operator $A.$ It is the smallest closed extension of $A$ in the sense that if $A\subset B$ and $B$ is closed, then $\bar{A}\subset B.$

Could anyone help with a proof? (I can guess this, but not sure what to prove...)

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  • $\begingroup$ It is the definition of closure of an operator $\endgroup$ – sinbadh Jan 2 '16 at 8:44
  • $\begingroup$ How to define the closure of operator, could you please express that mathematically? $\endgroup$ – math101 Jan 2 '16 at 9:36
  • $\begingroup$ As the smallest closed extension in the sense that you wrote. How do you define it? $\endgroup$ – sinbadh Jan 2 '16 at 9:46
  • $\begingroup$ You could also take the closure of the graph of $A $ in $X \times Y $, where $A:X'\to Y $ and $X'\leq X $. If this closure is a graph of a linear operator, call this operator $\overline {A} $. $\endgroup$ – PhoemueX Jan 2 '16 at 11:46
  • $\begingroup$ you mean, the closure of $A$ can be defined as the smallest closed extension of $A$ in the sense that if $A\subset B$ and $B$ is closed? $\endgroup$ – math101 Jan 2 '16 at 11:46

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