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It is my understanding that when an operator is densely defined, then the adjoint operator exists. But, if $X,\, Y$ Banach spaces and $A:D(A)\subset X\to Y$ linear operator.

Why is it necessary for $D(A)$ to be dense in $X$ to have an existence of the adjoint $A^*$?

If $D(A)$ is not dense in $X$, not necessarily the adjoint exists, right?

eraldcoil
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    If $D(A)$ is not dense then the adjoint is not uniquely defined. – Kavi Rama Murthy Nov 03 '21 at 05:49
  • The adjoint holds $(A^u,v)=(u,Av),\quad v\in D(A).$ If exists another $B^$ such that $( B^u,v)=(u,Av),\quad v\in D(A)$ then $(A^u-B^*u,v)=0,\quad v\in D(A).$

    How can use the dense property of $D(A)$ To demonstrate that $A^=B^$?

     – eraldcoil Nov 03 '21 at 05:53
  • $(A^{}u-B^{u},v)=0$ for all $v \in D(A)$ implies that $(A^{}u-B^{u},v)=0$ for all $v \in X$ which implies that $A^{}u-B^{}u$ is orthogonal to itself Hence it is $0$. – Kavi Rama Murthy Nov 03 '21 at 05:57
  • I understood perfectly. $(A^u-B^u) \perp X$ then $A^u-B^u=0$. But one last question arose. The adjoint is $A^:D(A^)\subset Y^\to X^$, $B^:D(B^)\subset Y^\to X^$. In principle, $u\in Y^$ (is a functional in $Y$) then $A^u,, B^*u$ are functional in $X$. What does it mean for a functional on $X$ to be perpendicular to $X$? – eraldcoil Nov 03 '21 at 06:03
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    $A^{}$ is a functional, $A^{}u$ is an element of $X$. – Kavi Rama Murthy Nov 03 '21 at 06:07
  • I think I have it. If $f\in X^*, f\perp x$ means $(f,x)=0$ for all $x\in X$. Equivalent is $f(x)=0$ for all $x\in X$. Then $f=0$. – eraldcoil Nov 03 '21 at 06:08
  • I had understood that $A^:D(A^)\subset Y^\to X^$... then $A^u\in X^$... – eraldcoil Nov 03 '21 at 07:05
  • Yes, I made a mistake. $A^{}u \in X^{}$. – Kavi Rama Murthy Nov 03 '21 at 07:19

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