Diagonal

The diagonal ${D=\{(x,x):0\le x\le 1\}}$ of the unit square ${Q=[0,1]\times [0,1]}$ is a mysterious thing. For one thing, its length is an irrational number. But this is not what I’m writing about.

The diagonal has zero area. Lebesgue integrable functions on ${Q}$ form the normed space ${L^1(Q)}$ which, upon closer inspection, consists not of functions but of their equivalence classes. Two functions ${f,g}$ are equivalent if the set ${\{f\ne g\}}$ has zero area. In particular, changing all the value of ${f}$ on the diagonal ${D}$ does not change ${f}$ as an element of ${L^1(Q)}$ . The logical conclusion is that given an element ${f\in L^1(Q)}$ , we have no way to give a meaning to the integral of ${f}$ over ${D}$ .

But wait a moment. If I take two square integrable function ${u,v\in L^2[0,1]}$ , then the expression ${\int_0^1 u(x)v(x)\,dx}$ makes perfect sense. On the other hand, it represents the integral of the function ${f(x,y)=u(x)v(y)}$ over the diagonal ${D}$ .

Pushing this further, if an element ${f\in L^1(Q)}$ can be represented as ${f(x,y)=\sum_{k=1}^n u_k(x)v_k(y)}$ for some ${u_k,v_k\in L^2[0,1]}$ , then ${\int_D f}$ is naturally defined as ${\sum_{k=1}^n \int_0^1 u_kv_k}$ . I can even take infinite sums, assuming that everything converges.

This is confusing. If I pick up an element of ${L^1(Q)}$ off the sidewalk, how will I know if it’s safe to integrate it over the diagonal? The existence or nonexistence of the decomposition into sum of products is not obvious.

I guess a satisfactory answer is given by the notion of a Lebesgue point. Given ${f\in L^1(Q)}$ and a point ${p}$ of ${Q}$ , consider the following statement:

$\displaystyle \exists y_0\in \mathbb R \text{ such that } \lim_{r\rightarrow 0} \frac{1}{r^2} \int_{|x-p|<r} |f - y_0| =0 \ \ \ \ \ (1)$

The validity of (1) and the value of ${y_0}$ are not affected by the choice of a representative of ${f}$ . If (1) holds, ${p}$ is called a Lebesgue point of ${f}$ , and we can think of ${y_0}$ as the “true” value of ${f(p)}$ (whether or not our representative agrees with that value). It’s a theorem that almost every point of the domain of ${f}$ is a Lebesgue point. The meaning of “almost every” corresponds to the measure under consideration: on the plane it’s the area, on a line it’s the length.

The product ${f(x,y)=u(x)v(y)}$ has a special feature. Since almost every ${x\in [0,1]}$ is a Lebesgue point of ${u}$ , and a.e. ${y\in[0,1]}$ is a Lebesgue point of ${v}$ , it follows that almost every point of the diagonal (in the sense of linear measure) is a Lebesgue point of the product ${f}$ . (It helps to integrate over small squares instead of disks in (1), which does not change anything.) This makes it possible to define ${\int_D f}$ unambigiously.

The sums of products also have the property that almost every point of ${D}$ is a Lebesgue point. And other elements of ${L^1(Q)}$ may also have this property: they are safe to integrate diagonally, too.

Diagonal

Related

Leave a Reply Cancel reply