1
$\begingroup$

Assume $f$ is a function from $\mathcal{R}^n$ to $\mathcal{R}$, $f$ is a convex and smooth function, can we express divergence theorem in this way?

$$\nabla\int\limits_{B_\delta}f(x+v)dv = \int\limits_{S_\delta}f(x+u)\frac{u}{\|u\|}du$$

where $B_\delta = \{y \mid \|y\|\leq \delta\}$ and $S_\delta = \{y \mid \|y\|= \delta\}$.

This is taken from equation (6.5) from page 109 of the book http://ocobook.cs.princeton.edu/OCObook.pdf Although it states that it is from Stokes theorem, but I believe it is actually divergence theorem.

Divergence theorem is usually used on the vector field, i.e., $\vec{f}$, can we generalize to the case of $f$ defined here? What about the dot product of $\langle\vec{f}, u\rangle$ in the original divergence theorem, can we simply state it in the above way? Why the gradient operator in the above equation is moved outside?

$\endgroup$
0
$\begingroup$

The notation is valid.

$$ \nabla \int_{B_{\delta}} f(x + \nu) d\nu $$

The gradient of the integral works vica versa due to the fundamental thereom of calculus. Both are good measures of how much of a function is contained within its limit, in this case topological Ball.

Integrating first gives the amount within the ball, and the derivative afterward takes into account its continuity. Deriving first considers its continuity and then integrates under that. You either have a volume of change or a change of volume, both as a representation of the original function $ f $.

$$ \int_{S_{\delta}} f(x + u) \frac{u}{||n||} du $$

This part is still valid as a measure of flux however think of it as a convolution of the function with respect to the added measure $ u $. Because the absolute value of $ u $ is added on the end as a vector, the integrand now represents the area of $ f $ that is under $ u $. I.e the protrusion of f with respect to $ u $.

It can still be stated as a dot proudct:

$$ \langle \partial_{\nu}, f(x+\nu) \rangle_{B_{\delta}} = \langle f(x+\nu), \frac{u}{||u||} \rangle_{S_{\delta}} $$

bracket notation stands for many things, one being an integral:

$$ \langle a,b \rangle_{x} = \int ab .dx $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.