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For a scalar function $g$, and a vector function $f$, $$ | \nabla ( (\nabla g) \cdot f ) | \leqslant |f| \cdot \text{Something} $$ Is this inequality possible? If possible, what would "$\text{Something}$" be?

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Here, assume that $g:R^3 \to R$ and $ f \in R^3 $ – Misaj Apr 22 '12 at 13:20
Language nitpick: "Nabla" is how you write the name of the symbol "$\nabla$", and is used to refer to it as a particular collection of pixels or ink. The name of the operation (or mathematical object) represented by that symbol in your formula is "del" or "gradient". – Henning Makholm Apr 22 '12 at 15:41
Ashuley, please be sure to accept an answer that has helped you. :) – tentaclenorm May 14 '12 at 15:48
up vote 0 down vote accepted

Here it goes an improvement of the last estimative i did before.

Assuming $F(x)=(f_1(x),f_2(x),f_3(x))$ a differentiable vector function of the variable $x=(x_1,x_2,x_3)$, lets calculate $\nabla ( (\nabla g) \cdot F )$. I will use the notation $\frac{\partial}{\partial x_i}g=g_{x_i}$. All the sums will be with $i=1,2,3$:

\begin{eqnarray} \nabla((\nabla g)\cdot F) &=& \nabla\left(\sum g_{x_i}f_i\right)\\ &=&\left(\left(\sum g_{x_i}f_i\right)_{x_1},\left(\sum g_{x_i}f_i\right)_{x_2},\left(\sum g_{x_i}f_i\right)_{x_3}\right)\\ &=&\sum\left(\begin{array}{c} g_{x_ix_1}f_i \\ g_{x_ix_2}f_i \\ g_{x_ix_3}f_i \end{array}\right)+\sum\left(\begin{array}{ccc} g_{x_1}(f_i)_{x_1} \\ g_{x_2}(f_i)_{x_2} \\ g_{x_3}(f_i)_{x_3} \end{array}\right)\\ &=&\textrm{D}^2g\cdot F+A \end{eqnarray}

where $A$ is the second term beside $\textrm{D}^2g\cdot f$. Lets take a more carefull look at $A$.

\begin{eqnarray} A&=&\left(\begin{array}{c} g_{x_1}[(f_1)_{x_1}+(f_2)_{x_1}+(f_3)_{x_1}] \\ g_{x_2}[(f_1)_{x_2}+(f_2)_{x_2}+(f_3)_{x_2}] \\ g_{x_3}[(f_1)_{x_3}+(f_2)_{x_3}+(f_3)_{x_3}] \end{array}\right)\\ &=&\left(\begin{array}{ccc} A_1 & 0 & 0 \\ 0 & A_2 & 0 \\ 0 & 0 & A_3 \end{array}\right)\cdot\nabla g \end{eqnarray}

where $$A_i=F_{x_i}\cdot(1,1,1)=\sqrt[3]{2}\cdot\left(F_{x_i}\cdot\frac{1}{\sqrt[3]{2}}(1,1,1)\right)=:\sqrt[3]{2}\cdot\left(F_{x_i}\cdot u\right)$$

where $u\in\mathbb{S}^2$ is a unitary vector. Hence,

$$|\nabla((\nabla g)\cdot F)|\leqslant\sup_{w\in\mathbb{S}^2}|\textrm{D}^2g \cdot w|\cdot|F|+|\nabla g|\cdot\max|A_i|$$

It is all indicating we need hypotesys on $g$ and its derivatives, as well as on $F$ and its derivatives (because $A_i$ depends on it) to get some useful inequality.

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I dont know exactly what kind of estimative you are looking for. Here it goes one: I assume $f\in\mathbb{R}^3$ is a constant vector, and change the notation $f$ by $v$. We see that

\begin{eqnarray} \nabla((\nabla g)\cdot v)&=&\nabla\frac{\partial g}{\partial v}\\ &=&\left(\frac{\partial}{\partial x}\frac{\partial g}{\partial v},\frac{\partial}{\partial y}\frac{\partial g}{\partial v},\frac{\partial}{\partial z}\frac{\partial g}{\partial v}\right)\\ &=&\left(\frac{\partial}{\partial v}\frac{\partial g}{\partial x},\frac{\partial}{\partial v}\frac{\partial g}{\partial y},\frac{\partial}{\partial v}\frac{\partial g}{\partial z}\right)\\ &=&\frac{\partial}{\partial v}(\nabla g)\\ &=&\textrm{D}^2g \cdot v \end{eqnarray}

and hence,

$$|\nabla\left((\nabla g)\cdot v\right)|\leqslant C\cdot|v|$$

where $C=\sup_{w\in\mathbb{S}^2}|\textrm{D}^2g \cdot w|$, or, in other words, $C$ is the spectral radius of $\textrm{D}^2g$, wich of course depends on the point where you are evaluating the inequality.

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