Ortogonality of two variable functions

[LagrangeEuler]

In functional analysis functions $f$ and $g$ are orthogonal on the interval $[a,b]$ if
\int^b_a f(x)g(x)dx=0
But what if we have functions of two variables $f(x,y)$ and $g(x,y)$ that are orthogonal on the interval $[a,b]$. Is there some definitions
\int^b_a f(x,z)g(z,y)dz=0?

 

[FactChecker]

Yes. The definition of "orthogonal" is very general. You can do integrals on both variables. As long as there is an "inner product" defined, the definition of "orthogonal" is also defined. And it is not hard to show that the double integral would have the properties of an inner product. (see http://mathworld.wolfram.com/InnerProduct.html )

 

[The Bill]

But what if we have functions of two variables $f(x,y)$ and $g(x,y)$ that are orthogonal on the interval $[a,b]$. Is there some definitions
\int^b_a f(x,z)g(z,y)dz=0?

The integral you've written is not what you seem to mean. First, it shouldn't be $dz$, it should be $dxdy$ or $dA$ for area. Also, it shouldn't be over a single interval the way you have it written. It should be over a set in the plane. That set could be a box defined by two intervals, one over $x$ and the other over $y$, or it could be any other arbitrarily shaped set, as long as it's measurable.

 

[Ssnow]

Hi, you must define the inner product on the space of function in two variables, because what you write doesn't make sense, in fact $\langle f,g\rangle=\int_{a}^{b}f(x,z)g(z,y)dz=F(x,y)$ that is a function of two variables $x,y$ and not a number (with $\langle\cdot,\cdot\rangle$ I denote the inner product) ...

Ssnow