Integral over boundary vs Integral over entire region (relationship?)

Question

Let ${\bf{x}}\in\mathbb{R}^2$ and $f({\bf{x}}): \mathbb{R}^2 \rightarrow (\epsilon,\infty)$, $\epsilon > 0$, that is, $f({\bf{x}})$ is scalar and positive for all $\bf{x}$. Let $S$ be bounded convex (polygon) subset of $\mathbb{R}^2$ and let $\partial S$ be the boundary of $S$ (so in $\mathbb{R}^2$ we have that $S$ is a non-smooth but piecewise continuous curve). Do I have any relationship between: $$ \int_{\partial S} f({\bf{x}}) d{\bf{x}}~~~~{\rm{and}}~~~~\int_{S} f({\bf{x}}) d{\bf{x}}$$

Perhaps, can I bound one using other, using the diameter of the polygon? I believe I can say $ \int_{\partial S} f({\bf{x}}) d{\bf{x}} \leq \int_{S} f({\bf{x}}) d{\bf{x}}$ but I am looking for something more tight.

I do not know if this helps, but this problem shows up if I want to evaluate how the mass of a Voronoi cell changes over time, where $S$ is the Voronoi cell and $f$ is the density function.

Answer 1

Assuming the function $f$ is bounded both below (by $\epsilon$ as specified) and above (by some maximum value $M$ which wasn't specified), the relationship is related to the Isoperimeteric Inequality. In two dimensions, this says that the sets perimeter is larger than its area,

$$ \left(\int_{\partial S} d{\bf x}\right)^2 \geq 4 \pi \int_S d{\bf x}. $$

Applying the bounds ($f > \epsilon$ and $f < M)$, this becomes,

\begin{align} \left(\int_{\partial S} f \, d{\bf x}\right)^2 & \geq \epsilon^2 \left(\int_{\partial S} d{\bf x}\right)^2 \\ & \geq 4 \pi \epsilon^2 \int_S d{\bf x}\\ & \geq \frac{4 \pi \epsilon^2}{M} \int_S f \, d{\bf x}. \end{align}

The reverse inequality cannot be established without additional assumptions: an arbitrarily thin rectangle can have a significant perimeter and an arbitrarily small area.

However Voronoi cells are convex and in many cases avoid problematic thinness as long as the Voronoi sites cannot get too close to each other. Let's proceed with two assumptions:

• $S$ is convex
• There is a constant $c>0$ such that some ball of radius $c\, {\rm diam}(S)$ is contained in $S$.

The second assumption is a requirement on the Voronoi cells that is roughly equivalent to good spacing of Voronoi generators or, equivalently, the aspect ratio of the Voronoi cells is bounded.

With convexity, the perimeter of a set can initially be bounded in terms of the area and diameter of set and then with the second assumption, becomes a bound on the area. The first bound (which seems to go back to Blaschke, W., Konvexe Bereiche gegebener konstanter Breite und kleinsten Inhalts., Math. Ann. 76, 504-513 (1915). ZBL45.0731.04.) says that the unit circle maximizes the perimeter over all convex sets with unit diameter. Scaling by the diameter, this is,

\begin{align} \left(\int_{\partial S} d{\bf x}\right)^2 & \leq \pi^2 \, {\rm diam} (S)^2\\ & \leq \frac{\pi}{c^2} \int_S d{\bf x}. \end{align}

A function $f$ bounded between $\epsilon$ and $M$ can be introduced into these integrals and passed through as before, \begin{align} \left(\int_{\partial S} f \, d{\bf x}\right)^2 & \leq \frac{M^2}{\epsilon}\cdot \frac{\pi}{c^2}\int_Sf \, d {\bf x}. \end{align}

Answer 2

The answer is basically: 'no'. There is no a priori relation between $$ \int_{\partial S} f({\bf{x}}) d{\bf{x}}~~~~{\rm{and}}~~~~\int_{S} f({\bf{x}}) d{\bf{x}}. $$ These quantities have to be considered completely independent. Asking for an estimate of the value of the first integral given the value of the second (or viceversa) is just as groundless as asking for the value $f(1)$ given the value $f(0)$ for a real function.

Instead, there is a relation between (for example) $$ \int_{\partial S} f({\bf{x}}) d{\bf{x}}~~~~{\rm{together\; with}}~~~~\int_{S} \nabla f({\bf{x}}) d{\bf{x}} $$ on one side and $$\int_{S} f({\bf{x}}) d{\bf{x}} $$ on the other side. This kind of relations are called "Poincaré inequalities"