Motivation: I am trying to understand the proof of Section 1.1 here p.4-5. I am trying to see the operator $-\Delta + \frac{a}{|x|^2}$ is positive semi-definite symmetric operator. (And so will have Friedrichs extension)
Let $\mathcal{L}_a^0$ denote the natural action of $- \Delta + \frac{a}{|x|^2}$ on $C_c^{\infty} (\mathbb R^{N}\setminus \{0\})$ ($a\geq - \left(\frac{N-2}{2} \right) ^{2}$).
Questions:
(1)How to show $$ \mathcal{L}_a^0 = \left( - \nabla + \sigma \frac{x}{|x|^2} \right) \cdot \left( \nabla + \sigma \frac{x}{|x|^2} \right)= - \Delta - \sigma (N-2 - \sigma) \frac{1}{|x|^2},$$ with $\sigma= \frac{N-2}{2}- \frac{1}{2} \sqrt{(N-2)^2 + 4a}$\ (2) How to use (1) to conculde that that for $\phi \in C_c^{\infty} (\mathbb R^N \setminus\{0\})$ we have $$ \langle \phi, \mathcal{L}_a^0 \phi \rangle = \int_{\mathbb R^N} |\nabla \phi(x) + \sigma \frac{x}{|x|^2} \phi(x)|^2 dx \geq 0 ?$$
My Rough Attempt: Let $N=3.$ (1) Note that\ $ \left( - \nabla + \sigma \frac{x}{|x|^2} \right) \cdot \left( \nabla + \sigma \frac{x}{|x|^2} \right)= \left( -\frac{\partial}{\partial x_1}+\sigma \frac{x_1}{|x|^2} , ..., -\frac{\partial}{\partial x_3}+\sigma \frac{x_1}{|x|^2} \right) \cdot \left( \frac{\partial}{\partial x_1}+\sigma \frac{x_1}{|x|^2} , ..., \frac{\partial}{\partial x_3}+\sigma \frac{x_1}{|x|^2} \right) = -\Delta + \sigma^2 \frac{1}{|x|^2 }$
I am wondering, in the aboe identity why one could get: $\mathcal{L}_a^0= -\Delta - \sigma (3-2-\sigma) \frac{1}{|x|^2} = - \Delta +(\sigma^2 - \sigma) \frac{1}{|x|^2}$
I have noticed that: $-\sigma + \sigma^2 =a$
(2) In proving, second how should I interprest inner product: should I take $\langle f, g\rangle = \int_{\mathbb R^3} f(x) \overline{g(x)} dx$ Or $\langle f, g\rangle = \int_{\mathbb R^3} \nabla f(x) \cdot \overline{\nabla g(x)} dx ?$
Is it true that: $|\nabla \phi \cdot \nabla (-\Delta - \sigma(N-2-\sigma)\frac{1}{|x|^2})\phi(x)|= |\nabla \phi(x) + \sigma \frac{x}{|x|^2} \phi(x)|$? (How should I proceed...)