Why don't we multiply generalized functions?

In summary, the conversation discusses the concept of generalized solutions of nonlinear PDEs and how they can be multiplied with functions from ##C^\infty(\mathbb{R})##. It is shown that this operation is well-defined and has certain properties, but it is also demonstrated that there cannot be a binary operation with values in ##\mathcal D'(\mathbb R)## that satisfies all standard properties of arithmetic multiplication.
  • #1
wrobel
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Because it drives to contradictions. Here is a nice example from E. Rosinger Generalized solutions of nonlinear PDE.

We can multiply generalized functions from ##\mathcal D'(\mathbb{R})## by functions from ##C^\infty(\mathbb{R})##. This operation is well defined. For example $$x\delta(x)=0\in \mathcal D'(\mathbb{R}),\quad x\cdot\frac{1}{x}=1\in \mathcal D'(\mathbb{R}),\quad \frac{1}{x}\in \mathcal D'(\mathbb{R}).$$
On the other hand ##C^\infty(\mathbb{R})\subset \mathcal D'(\mathbb{R})##

Ok then:)
$$\delta=\Big(x\cdot\frac{1}{x}\Big)\cdot\delta=\frac{1}{x}\cdot(x\delta)=0.$$
 
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  • #2
I have interpreted ##x \delta(x)=0## as for support of good behavior function, i.e.
[tex]\int f(x) x \delta(x) dx= 0 [/tex]
for f(x) such that f(0) is finite. 1/x does not satisfy it.
 
  • #3
This example shows that you can not define a binary operation of ##\mathcal D'(\mathbb R)\times \mathcal D'(\mathbb R)## with values in ##\mathcal D'(\mathbb R) ## such that
1) this operation acts on ##C^\infty(\mathbb{R})\times \mathcal D'(\mathbb R)## in the standard way;
2) this operation possesses the standard properties of the arithmetic multiplication
 

1. Why can't we multiply generalized functions?

Generalized functions, also known as distributions, are not actual functions but rather mathematical objects that generalize the concept of a function. They can have infinite or discontinuous values at certain points, making it impossible to define a multiplication operation between them.

2. What happens if we try to multiply generalized functions?

If we try to multiply generalized functions, we may end up with undefined or infinite values, which do not have any meaningful interpretation. This is because generalized functions are defined through their action on test functions, rather than being actual functions with well-defined values at each point.

3. Can we find a way to multiply generalized functions?

No, there is no consistent way to define multiplication between generalized functions. This is due to the fact that they are defined through their action on test functions, and not through a specific formula or set of values. Any attempt to define multiplication would lead to inconsistencies and contradictions.

4. Are there any alternative operations to multiplication for generalized functions?

Yes, there are alternative operations that can be defined for generalized functions, such as convolution and differentiation. These operations are consistent and well-defined, unlike multiplication, and are often used in applications involving generalized functions.

5. Why do we need to use generalized functions if we can't multiply them?

Generalized functions are powerful mathematical tools that allow us to solve problems involving distributions, such as differential equations and boundary value problems. While we may not be able to multiply them, their other properties and operations make them essential in many areas of mathematics and physics.

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