# Self-adjoint operator in the space of twice continuously differentiable functions

There is a problem in the textbook with which I am having difficulties.

Prove that operator $A$: $Ay=xy''+y'$ defined on space of twice continuously differentiable functions (scalar product is defined as $(y,z)=\int_1^3y(x)z(x)dx$) is self-adjoint operator. And there is the boundary conditions $y'(1)=y(1)$ and $y'(3)=0$.

I tried to do it by definition of self-adjoint operator $(Ay,z)=(y,Az)$.

1. $$(Ay,z)=(xy'',z)+(y',z)$$
2. The first term. $$\int_1^3xy''(x)z(x)dx=\int_1^3xz(x)dy'(x)=xz(x)y'(x)|_1^3-\int_1^3y'(x)\left(z(x)+xz'(x)\right)dx=$$ $$=xz(x)y'(x)|_1^3-\int_1^3\left(z(x)+xz'(x)\right)dy(x)=xz(x)y'(x)|_1^3-(xz'(x)+z(x))y(x)|_1^3+$$ $$+\int_1^3y(x)\left(2z(x)'+xz''(x)\right)dx$$
3. The second term. $$\int_1^3y'(x)z(x)dx=\int_1^3z(x)dy(x)=z(x)y(x)|_1^3-\int_1^3y(x)z'(x)dx$$
4. $$(Ay,z)=(xy'',z)+(y',z)=\{xz(x)y'(x)|_1^3-(xz'(x)+z(x))y(x)|_1^3+z(x)y(x)|_1^3\}+\int_1^3y(x)\left(z(x)'+xz''(x)\right)dx=$$ $$=\{xz(x)y'(x)|_1^3-(xz'(x)+z(x))y(x)|_1^3+z(x)y(x)|_1^3\}+(y,Az)=x\{z(x)y'(x)|_1^3-z'(x)y(x)|_1^3\}+(y,Az)$$

It is clear that the first term must vanish, but there is $z(1)$,$z(3)$ and $z'(1),z'(3)$ and I don't know what I should do with them.

Maybe I'm missing something or I'm wrong somewhere or there is some other ways to prove it?

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First, observe that $$Ay(x)=xy''(x)+y'(x)=(xy'(x))'.$$ Then, by integration by parts and given $y'(3)=0$: $$(Ay,z)=\int_1^3 (xy'(x))'z(x)dx=y'(1)z(1)-\int_1^3xy'(x)z'(x)dx.$$ On the other hand, still by integration by parts and using $z'(3)=0$: $$(y,Az)=\int_1^3 y(x)(xz'(x))'dx=y(1)z'(1)-\int_1^3y'(x)xz'(x)dx.$$ Since $y(1)=y'(1)$ and $z(1)=z'(1)$, is is now clear that $(Ay,z)=(y,Az)$ for every $y,z$ in the space under study.