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Rudin asked:

Let $K$ be a triangle (two dimensional figure) in the plane, let $H$ be the set consisting of the vertices of $K$, and let $A$ be the set of all real functions $f$ on $K$, of the form $$f(x,y)=\alpha x+\beta y+\gamma,\alpha,\beta,\gamma\in \mathbb{R}$$

Show that to each $(x_{0},y_{0})\in K$ there corresponds a unique measure $\mu$ on $H$ such that $$f(x_{0},y_{0})=\int_{H}fd\mu$$

I am slightly at lost with what Rudin means here by "vertices of $K$", does he just mean points in $K$? Or just the boundary of $K$? Suppose he mean the points in the whole triangle, I also do not know how to show the uniqueness (the existence is trivial by section 5.22).

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he means the three endpoints of the sides of the triangle. so the measure in question is on a finite measure space! –  user29743 Dec 30 '12 at 21:03
    
:(. Then 5.22 would not work as I need $H$ to be the boundary of $K$, but now $H$ is the boundary of boundary of $K$.... –  Bombyx mori Dec 30 '12 at 21:04
    
the last thing I said made no sense! sorry. –  user29743 Dec 30 '12 at 21:06
    
@countinghaus: No, I think I was wrong - you are right. Since the maximum of $f$ is in the 3 vertices, this would actually work. Thanks! I see why it is unique but I do not know how to prove it. –  Bombyx mori Dec 30 '12 at 21:09

1 Answer 1

up vote 4 down vote accepted

The vertices are the corners of the triangle. Suppose $H = \{v_1, v_2, v_3\}$, and assume that the points are not collinear (otherwise the uniqueness conclusion is false).

Then $K = \text{co} \{v_i\}_i$, and hence $x \in K$ iff $x = \sum_i \lambda_i v_k$ with $\sum_i \lambda_i = 1$ and $\lambda_i \geq 0$. Furthermore, since the points of $H$ is not collinear, the $\lambda_i$ are unique. The $\lambda_i$ are known as the barycentric coordinates of $x$ with respect to $\{v_i\}_i$.

Also, note that each $f \in A$ is affine, that is, if $\sum \lambda_i = 1$, then $\sum_i \lambda_i f(v_i) = f(\sum_i \lambda_i v_i)$.

To show existence, define the measure $\mu_x A = \sum_i \lambda_i 1_A(v_i)$, where the $\lambda_i$ are the barycentric coordinates of $x$. Then if $f \in A$, we have $\int f d \mu_x = \sum_i \lambda_i f(v_i) = f(\sum_i \lambda_i v_i) = f(x)$.

To show uniqueness, let $x \in K$ and suppose $\mu$ is a measure supported on $H$ such that $f(x) = \int_H f d \mu$ for all $f \in A$. This means $\int_H f d \mu = \int_H f d \mu_x$ for all $f \in A$. Since $\mu$ is supported on $H$, it has the form $\mu A = \sum_i \mu\{v_i\} 1_A(v_i)$, so we now have $\sum_i \lambda_i f(v_i) = \sum_i \mu \{v_i\} f(v_i)$, for all $f \in A$. By choosing $f_k(x) = x_k$ (ie, the coordinate functions), we see that this implies $\sum_i \lambda_i v_i = \sum_i \mu \{v_i\} v_i$, and by uniqueness of the barycentric coordinates, we have $\mu \{v_i\} = \lambda_i = \mu_x \{v_i\}$ for all $i$, hence we have uniqueness.

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