Let $\varphi \in C_0^\infty$, $\varphi \neq 0$.
We'll consider the inner product in $L^2.$
Let $\alpha ,\beta$ be multi-indices, $m\in \mathbb{N}$ such that $|\alpha|,|\beta|\leq m$ and set$$\varphi _{\alpha \beta}=\left \langle \dfrac{D^\alpha}{\alpha !},\dfrac{D^\beta}{\beta !}\right \rangle =\dfrac{1}{\alpha !\beta !}\int D^{\alpha}\varphi \overline{D^\beta \varphi}\,dx.$$Now consider $t=(t_\alpha )$ a "array" of complex numbers, such that $t_\alpha =t_{\alpha '}$, where $\alpha '$ is a permutation of $\alpha$.
Define $u(x)=\sum \limits _{|\alpha |\leq m}t_\alpha D^\alpha \varphi(x)$.
Define the quadratic form in $t$ defined by$$Q(t)=Q((t_\alpha ))=\sum \limits _{i\mid \alpha _i\leq m}\sum \limits _{i\mid \beta \leq m}t_\alpha \overline{t_\beta}\varphi _{\alpha \beta}.$$How to prove that in this conditions, holds$$\sum \limits _{|\alpha |\leq m}\sum \limits _{|\beta |\leq m}t_\alpha \overline{t_\beta}\varphi _{\alpha \beta}=\int \left |\sum \limits _{|\alpha|\leq m}\frac{t_\alpha D_\alpha \varphi}{|\alpha | !}\right |^2\,dx\hspace{2mm}?$$I saw this in Hormander's Thesis, this is Theorem 2.2 in page 179. He just say that this equality holds, but I can't see that.
I dont need to use that fact, I just want to prove that this quadratic form is positive definite to use coerciveness.
Can anybody help me, please?
P.S: I know that the matrix $(\varphi _{\alpha \beta})$ is hermitian, normal and diagonalizable. But I couldn't prove that the eigenvalues are strictly positive.