I have a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ representing a surface $z=f(x,y)$ in euclidean space. Is there an elegant way to apply the chain rule to expand $\nabla||\nabla f||_2$, where $||\cdot||_2$ is the euclidean norm?
Gradient of norm of gradient of scalar field
vector analysisvector-spaces
Best Answer
Kind of. But we want to do it very rigorously (at least one time). Furthermore we will assume that $f \in C^2(\mathbb{R}^2)$, i.e. that it is twice continuously differentiable. You will later see why.
Now let $g(x) := \lVert x \rVert_2$, $g:\mathbb{R}^2 \rightarrow \mathbb{R}$. Note that we have $$ \partial_{x_1} g(x) = \partial_{x_1} \sqrt{x_1^2 + x_2^2} = \frac{x_1}{\sqrt{x_1^2 + x_2^2}} $$ by the chain rule for one variable. You can do the same computation to get $\partial_{x_2} g$. So, (we set $Dg := \nabla g^\top$) $$ \nabla g(x) = \frac{x}{g(x)} \implies Dg(x) = \frac{x^\top}{g(x)} $$ which is a formula that can be very useful if you know it by heart. But this also means that $g \in C^1(\mathbb{R} \setminus \lbrace (0, 0)^\top \rbrace)$.
Furthermore let $h: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $, $h(x) := \nabla f(x)$. Note that $Dh(x) = Hf (x)$. So, for $x \in Z$ where $Z := \lbrace y \in \mathbb{R}^2: \nabla f(x) \neq 0 \rbrace $: $$ D (\lVert \nabla f(x) \rVert_2) = D(g \circ h)(x) = Dg(y)\vert_{y = h(x)}Dh(x) = \frac{1}{\lVert \nabla f(x) \rVert_2}\nabla f(x)^\top Hf(x) $$ We conclude: $$ \nabla \lVert \nabla f(x) \rVert_2 = \left( \frac{1}{\lVert \nabla f(x) \rVert_2}\nabla f(x)^\top Hf(x) \right)^\top = \frac{1}{\lVert \nabla f(x) \rVert}_2 Hf(x) \nabla f(x) $$ I think the latter is the expression you looked for.
Some nice facts: The assumption $f \in C^2(\mathbb{R})$ allowed us to write $Hf(x)^\top = Hf(x)$ in the very last step. This is calles Schwartz's theorem. Also, the set $Z$ is open: We can write $$ Z = \nabla f^{-1}((-\infty, 0) \cup (0, \infty)). $$ So $Z$ is the preimage of an open set under a continuous function, hence open. (The topology is always the euclidian one...)