# Clarification on Continuous Matrices and reference request

I have not been able to obtain a clear definition of what is currently called a 'continuous matrix' nor 'continuous matrix operators.'

It is unclear if the definition would involve a matrix with continuous elements such as continuous functions, or if somehow the matrix itself is to be considered continuous. In the latter case, I would imagine an example of a continuous matrix to be the Green's Function for a function, where integration of a function by its Green's function would be akin to some sort of continuous matrix multiplication.

So I would like to ask:

1. What is the current accepted definition for a continuous matrix and a continuous matrix operator?

2. Is there an existing definition for a continuous matrix, of the type I mention, where Green's Function is an example of such a matrix?

3. If there is such a definition (2.) then does someone know of a clear reference for the theory? I would appreciate both pure and applied works.

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I've never seen the term used anywhere, and a Google search returns very sporadic results, so I don't think this is even an accepted term in the mathematical community. Where in the world are you getting it from? The closest concepts I've found are the product and factorization of matrices that have components which are functions. –  anon Jul 8 '11 at 3:10
I believe it's somtimes used in the sense that the coefficients denpend continuously on a parameter - like in the book I link. Nevertheless, it would be better if @Roberto would provide some context. books.google.com/… –  Martin Sleziak Jul 8 '11 at 5:48
Assuming that you matrix depends on some parameter t it could also mean that the mapping $t \to A(t)$ is continuous in various operator topologies. For example in the uniform topology this means that for every $\epsilon > 0$ there exists $\delta > 0$ such that $\|A(t_1) - A(t_2)\| < \epsilon$ whenever $|t_1 - t_2| < \delta$. –  user12014 Jul 8 '11 at 6:32
Thank you for your responses. If a definition was not readily available then either the theory doesn't exist or is not widely disseminated, which answers my question. –  Roberto Miguez Aug 30 '11 at 3:01