# Cauchy-Schwarz inequality on double-summation term

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?

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 $j$.

• Thanks for your answer (which I agree with). According to my book, it should equal $|v|^2 |\nabla v^|$, but that can't be right? Mar 1, 2014 at 11:55
• @BillyJean: It depends on what 'it' is. I'll edit my answer to give some more insight; we can play with $w$. Mar 1, 2014 at 12:01
• @BillyJean: No edit, just a comment due to time constraints: If we can show that $\vert w \vert \leq \vert v \vert \vert \nabla v \vert$, then we are done. It's not clear what's meant by $\nabla v$ since both are vectors. Is it a scalar? Is it a matrix? For a matrix $A$, we have $\vert Av \vert \leq \Vert A \Vert \vert v \vert$, where $\Vert A \Vert$ denotes the induced matrix norm of $A$. Mar 1, 2014 at 12:11
• thanks, I will give it a shot and report back Mar 1, 2014 at 12:13