0

Let's consider linear operators on the set of complex-valued functions to the same set. I wonder to which categories such operators can be classified. All linear operators I encountered so far fall into the following categories:

  1. Differintegral of complex or variable order $L:f\to\mathbb{D}^{g}f$

  2. Multiplication by a variable or constant coefficient $L:f\to g f$

  3. Right composition $L:f\to f\circ g$

  4. Finite differences of complex or variable order $L:f\to\Delta^{g}f$

  5. Convolution with a function $L:f\to f * g$

  6. Successive combination of finite number of the operators belonging to the above classes

  7. Sum of finite number of the operators belonging to the above classes

For example, Fourier transform can be expresses through combination of convolution and some other operators from the list.

I am also aware about limits, but they would differ from the abovementioned only in finite amount of points.

So my question is: are there categories of linear operators on the same set that do not belong to the mentioned categories and differ from them more than just in countable number of points?

Are there any examples of such operators?

Anixx
  • 9,119
  • 1
    Limits in what sense? If you are considering all functions, limits will surely not be defined in all cases... – Mariano Suárez-Álvarez Jan 12 '11 at 20:35
  • Since you are calling differentiation a linear operator on the set, I guess you are only considering the space of entire functions? It would be nice if you made this explicit, and also explicitly stated the domain. Also, unequal linear functions on a complex vector space cannot differ on just a countable set. You may want to start by finding the size of the set of linear transformations by considering a Hamel basis. – Jonas Meyer Jan 12 '11 at 21:58
  • "Also, unequal linear functions on a complex vector space cannot differ on just a countable set." - I meant contable set of points x, arguments of the function, not on contable set of functions. – Anixx Jan 12 '11 at 22:22
  • I already stated the domain: any complex-valued functions on the complex set. Are there any linear operators on that set of subset of it which do not belong to the mentioned categories? Any examples? – Anixx Jan 12 '11 at 22:24
  • @Anixx: Your first comment is inconsistent with your question. If your operators are defined on a space of functions, then the points in the domain of the operators are functions. You did not state the domain of the complex-valued functions, but now you say they are defined on "the complex set", which I guess means the set of complex numbers. I meant in my comment that it would be nice to include that information, along with the assumption that the functions are differentiable and any other important assumptions, explicitly in the question. – Jonas Meyer Jan 12 '11 at 22:48
  • Re-read the question, the domain of complex valued functions on complex numbers. What's missing? – Anixx Jan 12 '11 at 23:28
  • There is no assumption that the functions are differentiable, if there is an operator on non-differentiable functions that does not belong to the mentioned, but linear, I would be interested. – Anixx Jan 12 '11 at 23:32

1 Answers1

3

If I understand correctly, your vector space is $\mathbb{C}^{\mathbb{C}}$, the space of all functions $f: \mathbb{C} \rightarrow \mathbb{C}$. As a $\mathbb{C}$-vector space this has dimension $c^c = 2^c$. You have also written down $2^c$ different linear operators.

However, the space $\operatorname{End} \mathbb{C}^{\mathbb{C}}$ has dimension at least $2^{2^c}$. Thus you are incredibly far from having written down all the linear operators. To get the right cardinal number, consider all possible permutations of a basis of $\mathbb{C}^{\mathbb{C}}$.

Pete L. Clark
  • 97,892
  • But can all the linear operators on this space be counted by just describing or counting their general forms in principle, just like linear functions can? – Anixx Jan 28 '11 at 20:42
  • Also, even if they cannot, what are the other important categories? – Anixx Jan 28 '11 at 20:46
  • @Pete: $\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_02^{\aleph_0}} = 2^{2^{\aleph_0}} = 2^{\mathfrak{c}}$. See also http://math.stackexchange.com/questions/17914/cardinality-of-the-set-of-all-real-functions-of-real-variable/17915#17915 – Arturo Magidin Jan 28 '11 at 20:46
  • @Arturo: like I said, blanking. Cardinal exponentiation is not my strong suit, apparently. – Pete L. Clark Jan 28 '11 at 22:15
  • 1
    @Pete: We all have those moments... (-: – Arturo Magidin Jan 28 '11 at 22:17
  • @Anixx: Sorry, I don't really understand your followup questions well enough to answer them. If you have a further specific question, feel free to edit it into your post. – Pete L. Clark Jan 28 '11 at 22:18