functional analysis - Using Cauchy Schwarz for functions in Sobolev Space

Hi I just want to confirm something simple and check that the following is allowed:

The Cauchy-Schwarz inequality states if $A = ((a_{ij}))$ is a symmetric, non-negative $n \times n$ matrix then

$|\sum_{i,j=1}^{n}a_{ij}x_{i}y_{j}| \leq (\sum_{i,j=1}^{n}a_{ij}x_{i}x_{j})^{\frac{1}{2}}(\sum_{i,j=1}^{n}a_{ij}y_{i}y_{j})^{\frac{1}{2}}$ for $x,y \in \mathbb{R}^{n}$.

Instead of $x,y \in \mathbb{R}^{n}$ consider $u,v \in H_{0}^{1}(U) = W_{0}^{1,2}(U)$ (Sobolev Space).

In other words:

$|\sum_{i,j=1}^{n}a_{ij}u_{x_{i}}v_{x_{j}}| \leq (\sum_{i,j=1}^{n}a_{ij}u_{x_{i}}u_{x_{j}})^{\frac{1}{2}}(\sum_{i,j=1}^{n}a_{ij}v_{x_{i}}v_{x_{j}})^{\frac{1}{2}}$ for $u,v \in H_{0}^{1}(U)$ where $U \subset \mathbb{R}^{n}$ and where $u_{x_{i}}:= \frac{\partial u}{\partial x_{i}}$.

It seems that this would be fine for each $x \in U$ since $(u_{x_{i}}(x))_{i}$ and $(v_{x_{j}}(x))_{j}$ are just sequences of numbers. What do you think?