# Math Help - The adjoint of unbounded operator

1. ## The adjoint of unbounded operator

Hello. I'm trying to find the adjoint of A, where A is unbounded operator in $L^2[0,1]$, D(A)=C[0,1], (Af)t=f(0).
I think, $A^*$ is operator $A^*g=g$. Is this correct? And what is $D(A^*)$?

2. Originally Posted by karkusha
Hello. I'm trying to find the adjoint of A, where A is unbounded operator in $L^2[0,1]$, D(A)=C[0,1], (Af)t=f(0).
I think, $A^*$ is operator $A^*g=g$. Is this correct? And what is $D(A^*)$?
If $g\in D(A^*)$, with $A^*(g) = h$, then $\langle Af,g\rangle = \langle f,h\rangle$ for all f in C[0,1]. Writing the inner products as integrals, $\displaystyle\int_0^1\!\!\!f(0)\overline{g(t)}\,dt = \int_0^1\!\!\! f(t)\overline{h(t)}\,dt$. In particular, if f(0) = 0 then $\displaystyle\int_0^1\!\!\! f(t)\overline{h(t)}\,dt = 0$. But continuous functions vanishing at 0 are dense in $L^2[0,1]$, so if h is orthogonal to all such functions then h = 0.

So I reckon that $A^*$ is the zero operator. Next, what is the domain of $A^*$? Well, if $g\in D(A^*)$ and $A^*(g) = 0$ then $\displaystyle\int_0^1\!\!\!f(0)\overline{g(t)}\,dt = 0$ for all f in C[0,1]. That condition is equivalent to $\displaystyle\int_0^1\!\!\!g(t)\,dt = 0$. So it looks as though the domain of $A^*$ is $\{g\in L^2[0,1] : \langle g,1\rangle = 0\}$ (where 1 denotes the constant function 1 on [0,1]).

3. Yes, you are right! Thank you very much!