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Let $$F, G: \mathbb{R}\rightarrow \mathbb{R},$$ be two real smooth functions such that $$G(x)=\int_{-\infty}^{\infty}H(y)F(x-y)dy,$$ for a complex function $$H:\mathbb{R}\rightarrow\mathbb{C}.$$

Suppose now that $F(x)\ge 0$ and $G(x)\ge 0$ for all real $x$. Does it imply that $H(x)\in \mathbb{R}_{\ge 0}$ for all $x$?

It seems that the answer is positive, but I can't prove it.

Regards

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  • $\begingroup$ Is $G$ real or complex valued? $\endgroup$
    – pancini
    Oct 2, 2020 at 9:02
  • $\begingroup$ @ElliotG He/She mentioned $G(x)\geq 0$ which is a subset of real set. $\endgroup$
    – ConvXET
    Oct 2, 2020 at 9:35

1 Answer 1

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That is not true for every class of functions $H$. The reason is that, due to smoothness of $F$, we have \begin{equation} \int H(y) F(x-y) {\rm d} y = \int \tilde{H}(y) F(x-y) {\rm d} y , \end{equation} where $H(x)$ and $\tilde{H}(x)$ are different only in a zero Lebesgue measure set. Hence, we simply can make the value of $H(x)$ negative without changing $G(x)$.

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