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Let $T \in \mathscr D^\prime(\mathbb R)$ be a distribution and let $\delta_0$ be the usual Dirac mass at $0$ (in $\mathbb R$).

Let us define the functional $S:=T\otimes \delta_0$ in the following way: for every $\phi \in C_c^\infty(\mathbb R^2)$ we set $$ \langle S, \phi \rangle := \langle T, \phi(\cdot, 0) \rangle, $$ which makes sense being $\phi(\cdot, 0) \in C_c^\infty(\mathbb R)$.

Question. Is $S$ a distribution?

I do think the answer is yes, because the map $S$ is clearly linear, and continuity should also be true (?). Is this an instance of the so-called tensor product of distributions? To which distributions can this construction be generalized?

I have always heard that defining product of distributions is "hard" (apart in the trivial case, i.e. when one distribution is represented by a smooth compactly supported function).

Thanks.

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    $\begingroup$ You might want to have a look at Klainerman's notes. The chapter on distributions is a concise and well-done exposition of some more advanced (and interesting) things of distribution theory, such as the tensor product you mention in this question. Such things occupy many more pages in standard books. $\endgroup$ – Giuseppe Negro Oct 19 '16 at 20:50
  • $\begingroup$ the product of distributions is hard (and impossible in general), but not the direct product or the tensor tensor $\endgroup$ – reuns Oct 19 '16 at 21:08
  • $\begingroup$ @GiuseppeNegro Sorry for the delay in the answer. Thank you very much for the good reference, really interesting! $\endgroup$ – Romeo Oct 22 '16 at 18:27
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To prove continuity of $\phi$, it's enough to note that if $\phi_n\rightarrow\phi$ in $C_c^{\infty}(\mathbb{R}^2)$ then $$\psi_n=\phi_n(\cdot,0)\rightarrow\phi(\cdot,0)=\psi$$ in $C_c^{\infty}(\mathbb{R})$, which should follow from definitions (the derivatives of $\psi_n$ are partial derivatives of $\phi_n$).

The tensor product $T_1\otimes T_2$ of two distributions $T_1$ and $T_2$ is defined the same way as: $$\langle T_1\otimes T_2,\phi\rangle=\langle T_1, \langle T_2, \phi(x_1,\cdot)\rangle\rangle=\langle T_2, \langle T_1, \phi(\cdot,x_2)\rangle\rangle$$ It is the unique distribution which satisfies $$\langle T_1\otimes T_2,\phi_1\otimes\phi_2\rangle=\langle T_1,\phi_1\rangle\cdot\langle T_2,\phi_2\rangle$$ where $\phi_1\otimes\phi_2$ is the usual tensor product of functions: $$\big(\phi_1\otimes\phi_2\big)(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)$$

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  • $\begingroup$ I don't like your 1st line. You should use the semi-norms definition of continuity : $\|\psi\|_{k,m} \le\|\phi\|_{k,m}$ and if $\phi$ is compactly supported then $\psi$ is compactly supported too $\endgroup$ – reuns Oct 19 '16 at 21:11
  • $\begingroup$ I agree with the other comment, I do not understand what you are doing in the first line. The rest of the answer is perfect, thanks. $\endgroup$ – Romeo Oct 22 '16 at 18:27

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