# a proof about closable operator

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...)

• It is the definition of closure of an operator – sinbadh Jan 2 '16 at 8:44
• How to define the closure of operator, could you please express that mathematically? – math101 Jan 2 '16 at 9:36
• As the smallest closed extension in the sense that you wrote. How do you define it? – sinbadh Jan 2 '16 at 9:46
• 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 $\overline {A}$. – PhoemueX Jan 2 '16 at 11:46
• 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? – math101 Jan 2 '16 at 11:46