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.

Behold the mystery
Behold the mystery

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.

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