Cauchy-Schwarz inequality on double-summation term

Question

I have the following, where $v$ is a vector $$ v\cdot (v\cdot \nabla)v $$ which in index notation becomes $v_jv_id_iv_j$. I want to apply the Cauchy-Schwarz inequality on this, which is given by $$ |\sum_nA_nB_n|^2 \leq \sum_nA_n^2\sum_mB_m^2 $$ How do I use this (which only has a single summation index on the LHS) on $v_jv_id_iv_j$, which has two indices?

Answer 1

Assuming $\nabla$ is not just some vector, but a differential operator, and $v$ is a function, and for simplicity $n=3$, there are two vectors involved: $$v=\begin{pmatrix} v_1 \\v_2\\v_3\end{pmatrix} \text{ and } w := \begin{pmatrix} (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_1 \\ (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_2\\ (v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_3\end{pmatrix}.$$

Applying Cauchy-Schwarz to their scalar product yields

$$ \vert v \cdot (v \cdot \nabla)v \vert = \vert v \cdot w \vert\leq \Vert v \Vert \Vert w \Vert = \left(\sum_j v_j^2\right)^{1/2}\left(\sum_j ((v_1 \partial_x + v_2 \partial_y + v_3 \partial_z )v_j)^2\right)^{1/2}.$$

I'm deliberately not using a double summation here, but if you wanted to, you could reintroduce it to shorthand the rows of $w$, i.e. the second sum above.

Edit: Loosely speaking, keep the summation over $i$ attached to the second $v_j$, and do Cauchy-Schwarz for $j$.