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Is there some function $g: (0,+\infty) \to \mathbb{R}$ satisfying $$(1*g)(t)=t^{-1/2}$$ for all $t>0$? Here the sign $*$ represents the convolution. I tried apply the Laplace transform both site of the equality, but i arrive that $$g(t)=[\mathcal{L}^{-1}(\lambda^{\frac{1}{2}})](t)$$ so I can't develop it. Can somebody help me?

Jarbas Dantas Silva
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  • $$(1*g)(t)=\int_0^t g(s)ds=t^{-1/2}$$ For any smooth function $g(s)$ at $t\to0,$ LHS $\to0$ and RHS $\to\infty$, so such smooth function does not exist. – Svyatoslav May 28 '22 at 05:03
  • @Svyatoslav it dont need to be smooth. May be a $L^1_{loc}$ function for example. – Jarbas Dantas Silva May 28 '22 at 05:20
  • https://mathoverflow.net/questions/138137/does-the-inverse-laplace-transform-of-the-square-root-exist – Svyatoslav May 28 '22 at 05:27
  • I think that the idea is do not use laplace transform. – Jarbas Dantas Silva May 28 '22 at 16:55
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    $g$ exists as a distribution. If $$(t_+^{-3/2}, \phi) = \int_{\mathbb R^+} \frac {\phi(t) - \phi(0)} {t^{3/2}} dt,$$ then $\mathcal Lt_+^{-3/2} = -2 \sqrt {\pi s}$ and $H * (-t_+^{-3/2}/2) = H(t)/\sqrt t$ (the convolution is always well-defined for distributions with supports bounded on the same side). – Maxim May 30 '22 at 20:56
  • @Maxim I do not understant how define a laplace transform of a distribution. For example, what wold be $\int_0^\infty e^{-t\lambda} (t_+^{-3/2})dt$? – Jarbas Dantas Silva Sep 03 '22 at 22:44

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