Why don't we multiply generalized functions?

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  • #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
anuttarasammyak
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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
wrobel
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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
 

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