According to the Sturm-Liouville theorem, for any continuous function $p\in\mathcal{C}^0([0,1],\mathbb{R})$, there is a Hilbert basis (normlised) $(\psi_n)_{n\geq1}$ of $L^2([0,1],\mathbb{R})$ such that $\psi_n\in\mathcal{C}^2([0,1],\mathbb{R})(\forall n)$ and
$$-\psi_n''+p\psi_n=\lambda_n\psi_n$$
on $[0,1]$ and $\psi_n(0)=\psi_n(1)=0$. It's clear that $\psi_n\in H^1_0([0,1],\mathbb{R})$. If we give $H^1_0([0,1],\mathbb{R})$ a norm as $(u,v)_p=\int_0^1(u'v'+puv)(u,v\in H^1_0([0,1],\mathbb{R}))$, then $H^1_0([0,1],\mathbb{R})$ becomes a Hilbert space.
My question is whether the basis $(\psi_n)_n$ forms a basis for $(H^1_0([0,1],\mathbb{R}),(\cdot,\cdot)_p)$ under its inner product?
I tend to believe that it is right but I can not prove it. If each multiplied by some constant, then $(\psi_n)_n$ have norm $1$ and still remains orthogonal one to another (note as $\tilde{\psi_n}$). I want to show that for any element $u\in H^1_0([0,1],\mathbb{R})$, $\sum_{k=1}^{n}\tilde{\psi_k}(\tilde{\psi_k},u)_p$ converge in the norm $(\cdot,\cdot)_p$. But even this step, I have not yet succeeded.
Will someone be kind enough to give me some hints or give me some resources on this problem? Thanks in advance.
Best Answer
Let's show that $\psi_n$ is dense in $H_0^1$. Take $v\in H_0^1$ and suppose that $$\tag{1}(\psi_n,v)_p=(\psi_n',v')_2+(p\psi_n,v)_2=0,\ \forall\ n=1,...$$
Let's prove that $v=0$ which implies that the sequence $\psi_n$ is dense in $H_0^1$.
First, mutliply the equation $-\psi''_n+p\psi_n=\lambda_n\psi_n$ by $\psi_m$ and integrate by parts to conclude that
$$\tag {2} (\psi'_n,\psi'_m)_2=-(p\psi_n,\psi_m)_2, \ m\neq n $$
$$\tag {3}(\psi'_n,\psi'_n)_2=-(p\psi_n,\psi_n)_2+\lambda_n, \ m=n$$
Write $v=\sum_{i=1}^\infty (v,\psi_i)_2\psi_i$ and substitute it in $(1)$ to get
$$\tag{4} \sum_{i=1}^\infty (v,\psi_i)_2(\psi'_n,\psi'_i)_2+\sum_{i=1}^\infty (v,\psi_i)_2(p\psi_n,\psi_i)_2=0,\ \forall\ n=1,...$$
By combining $(2),(3)$ and $(4)$ we conclude that $$\tag{5} (v,\psi_n)_2\lambda_n=0,\ \forall\ n=1,...$$
From $(5)$ we conclude that $v=0$.
Remark: After some ponderation, I saw that $0$ can be a eigenvalue of this problem. This is because there is no assumptions of positivity of $p$, hence, I would like to point out that my argument only applies if $\lambda_n\neq 0$ for all $n$.