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Let $A:C^2[0,1]\to L^2[0,1]$ and $Af=f''$. Then we know that $A$ is an densely defined unbounded operator. How to determine $A^*$?

I know that this operator is not closed since it is not bounded: $||T(\sin(nx))||\geq ||\sin(nx)||$. Furthermore, it is not injective and surjective obviously.

When it comes to $A^*$, first, I want to compute $$D(A^*)=\{f\in L^2[0,1]:\exists c\geq 0 \text{ s.t. }\int_0^1fg''\leq c\left(\int_0^1|g|^2\right)^{\frac 1 2},\ \ \forall g\in C^[0,1]\}.$$ Here is where I am stucked. I think it may be raleted to integration by parts, but I don't know what to do next.

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If $f\in D(A^*)$, then the form $g\longmapsto \int_0^1 fg''$ is bounded, so by Riesz Representation Theorem there exists $f''\in L^2$ such that $$ \int_0^1 fg''=\int_0^1f''g. $$ So $f$ has a second weak derivative. And conversely, if $f$ has a second weak derivative, then $f\in D(A^*)$. So $$ \langle A^*f,g\rangle=\langle f,Ag\rangle=\int_0^1 fg''=\int_0^1f''g=\langle f'',g\rangle. $$ Thus $A^*$ is the operator that maps $f\longmapsto f''$ (in the weak sense),.

Martin Argerami
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