Showing that $Tp(x)=(1-x^2)p''(x)-2xp'(x) $ is self-adjoint

Question

Let $P_2(\Bbb{R})$ have the inner product $\langle p,q\rangle =\int\limits_{-1}^{1} p(x)q(x)dx$ and consider the operator $Tp(x)=(1-x^2)p''(x)-2xp'(x) = [(1-x^2)p'(x)]'$.

Show that T is self-adjoint.

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So far I have this:

$\langle p,T^*q\rangle =\langle Tp,q\rangle$

=$\langle(1-x^2)p''(x)-2xp'(x),q\rangle$

=$\int\limits_{-1}^{1} [(1-x^2)p'(x)]'q(x)dx$

Then I did integration by parts and got an answer that seems like it's not going to amount to anything. Is this a question which I should tackle by calculating the matrix of $T$ with respect to some basis and then taking the conjugate transpose to find $T^*$? If so, what basis should I use?

Answer 1

We have\begin{align}\bigl\langle T(p),q\bigr\rangle&=\int_{-1}^1\bigl((1-x^2)p'(x)\bigr)'q(x)\,\mathrm dx\\&=\bigl[\bigl((1-x^2)p'(x)\bigr)q(x)\bigr]_{x=-1}^{x=1}-\int_{-1}^1p'(x)(1-x^2)q(x)\,\mathrm dx\\&=-\int_{-1}^1p'(x)(1-x^2)q(x)\,\mathrm dx\\&=-\bigl[p(x)(1-x^2)q(x)\bigr]_{x=-1}^{x=1}+\int_{-1}^1p(x)\bigl((1-x^2)q(x)\bigr)'\,\mathrm dx\\&=0+\bigl\langle p,T(q)\bigr\rangle\\&=\bigl\langle p,T(q)\bigr\rangle.\end{align}Therefore, $T$ is self-adjoint.

Answer 2

$$Tp(x)=(1-x^2)p''(x)-2xp'(x) = [(1-x^2)p'(x)]'$$

then by integration by part we get, $$ \langle Tp,q\rangle = \int\limits_{-1}^{1} [(1-x^2)p'(x)]'q(x)dx \\= \underbrace{[(1-x^2)p'(x)q(x)]^1_{-1} }_{=0}- \int\limits_{-1}^{1} (1-x^2)p'(x)q'(x)dx \\=-\int\limits_{-1}^{1} [(1-x^2)q'(x)] p'(x)dx$$

doing integration by part again we get

$$ \langle Tp,q\rangle =-\int\limits_{-1}^{1} [(1-x^2)q'(x)] p'(x)dx\\=-\underbrace{[(1-x^2)p(x)q'(x)]^1_{-1} }_{=0} +\int\limits_{-1}^{1} [(1-x^2)q'(x)]' p(x)dx =\langle Tq,p\rangle $$

that is $$ \langle Tp,q\rangle =\langle Tq,p\rangle $$