regular tempered distributions

Asked Dec 27 '16 at 13:56

Active Dec 27 '16 at 19:46

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All regular* distributions of $\mathcal{D}'(\mathbb{R}^n)$ can be characterised as $L^1_{loc}(\mathbb{R}^n)$. Is there a similar characterisation of the regular distributions of $\mathcal{S}'(\mathbb{R}^n)$?

I often found, that all allgebraically bounded $f \in L^1_{loc}(\mathbb{R}^n)$, i.e. $\exists C>0, N>0:\ |f(t)|\le C(1+|t|)^N$, give regular tempered distributions, but for example this does not contain $\chi_{[-1,1]} \frac{1}{\sqrt{|t|}}\in L^1(\mathbb{R}^n)$, which should also be a regular tempered distribution? On the other hand, it should be "smaller" than $L^1_{loc}(\mathbb{R}^n)$ as $e^t$ won't give a regular tempered distribution.

(*) A tempered distribution $T\in \mathcal{S}'(\mathbb{R}^n)$ is called regular in this context, if it can be represented by an (ordinary) function $f\in L^1_{loc}(\mathbb{R}^n)$, i.e. $$\exists f\in L^1_{loc}(\mathbb{R}^n):\quad T(\phi)=\int_{\mathbb{R}} f \phi\ dx \ \qquad\forall \phi \in \mathcal{S}(\mathbb{R}^n)$$

edited Dec 27 '16 at 19:46

asked Dec 27 '16 at 13:56

julbern

What does it mean $\int f \phi dx$ when $f \not \in L^1_{loc}$ ? – reuns Dec 27 '16 at 18:25
I understand your question now, but the example $\chi_{[-1,1]} \frac{1}{t^2}$ is poorly chosen as it's not an integrable function on the line. Also, @user1952009's objection would be addressed by saying that $f$ is in $L_{loc}^1$ in the definition of a regular distribution. – Dec 27 '16 at 18:58
sorry my bad, I actually meant $\frac{1}{\sqrt{t}}$ and of course the set of regular tempered distributions is a subspace of $L^1_{loc}(\mathbb{R}^n)$ – julbern Dec 27 '16 at 19:09
Thanks for the link, I will read it now. – julbern Dec 27 '16 at 19:12
answer in: Which functions are tempered distributions? – julbern Dec 27 '16 at 19:30

regular tempered distributions

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