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I have a background in math, but I think the following construction appears often in physics.

Is there a term for a functional which maps one smooth function $f: R^n \rightarrow R$ to another $L(f):R \rightarrow R$ which represents the integral along each level set? That is, $L(f)(a) = \int_{f^{-1}(a)} d\mu $ where the measure $d\mu$ might be the induced measure from the metric on $R^n$, or it might also depend on the original $f$ in some intelligent way.

For instance, if the initial function is $f:R^2 \rightarrow R$ is defined $f(x)=x^2+y^2$ then $L(f)(a) = 2 \pi \sqrt{a}$.

If the original function is $f:R \rightarrow R$ is defined $f(x)=x^2$, then we'd have $L(f)(a)=0$ for $a<0$, $1$ for $a=0$, and $2$ for $a>0$. In both cases it might be more natural to integrate with a term derived from the value of $f'$ normal to the level set.

I suppose one could generalize to any smooth map $f:M \rightarrow M'$ from a Riemannian manifold to a another smooth manifold and get a new function that represents a computed volume of the preimage for each point: $L(f):M' \rightarrow R$.

My question: Is there a good way to compute $L(f)$ for some class of function $f$?

Any thoughts would be appreciated. It is reminiscent of (but I hope simpler than) stationary phase approximations and path integral calculations.

http://en.wikipedia.org/wiki/Stationary_phase_approximation

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If I remember correctly, what you are describing is close to the concept of a distribution function (it is the name that I am slightly unsure about).

Given a measure space $(X,\mathcal{E},\mu)$, and a measurable function $f:X\to \mathbb{R}$, a concept often used in classical and harmonic analysis is the notion of the distribution function $\lambda_f:\mathbb{R}_+\to\mathbb{R}$ defined as

$$ \lambda_f(a) = \mu\left( \{|f| < a\} \right) $$

that is, $\lambda_f$ is returns the measure of the set on which $|f|$ is bounded by $a$. One can reformulate the $L^p$ norms of $f$ in terms of the distribution function

$$ \|f\|_{L^p(X,\mu)} = p \int_0^\infty t^{p-1}\lambda_f(t) dt $$

a device that is very useful for analytical estimates. Now, formally speaking the functional $L(f)$ you wrote down is very similar to the derivative $\lambda_f'$.


Another possibility that you may want to look into is the notion of pull-back distributions. If $f:M\to\mathbb{R}$ is a smooth map with non-vanishing gradient, then given any distribution (generalized function) $k$ on $\mathbb{R}$ you can define the pull-back distribution $f_*k$ on $M$. The definition of $f_*k$ depends closely on the notion of $L(f)$ you wrote down (see e.g. the introduction to distributions by Friedlander and Joshi), and is a generalization of the co-area formula.


I am not exactly sure how to answer your second question of "is there a good way to compute". If you are asking whether given an analytic expression for the function $f$ whether you can have a "closed form" expression for $L(f)$, the answer is surely no, since the expression also depends on the chosen Riemannian metric/volume form/measure on $M$. If you are asking whether there are any convenient expressions to capture $L(f)$, I think you need to say a bit more about why you care about $L(f)$. You can certainly just write $L(f) = f_*\delta_a$ as the pull-back of the Dirac delta distribution on $\mathbb{R}$; whether that expression is a useful one for your purpose, I don't know.

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other way: $\lambda_{f}(a)=\mu(x:|f|>a)$ (Folland) –  TKM Feb 21 at 23:48

Using the Fourier transform representation of the Dirac measure on the level set, the volume of the level set (which is the required integral according to the examples given in the question) is given by (using the same notation as in the question):

$L(f)(a) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{d}k \,exp(-ika)\, \int\,|\nabla f|\,exp(ikf)\,\mathrm{d}\mu$

On a Riemannian manifold, the gradient is given by:

$\nabla f = \sum_{ij}g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$

The two examples given in the question can be worked out according to this formula (the second example very easily, the first one involves integrals of trigonometric functions of a squared angle, but it can also be solved analytically).

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