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$\DeclareMathOperator{\Mel}{M} \newcommand{\Rn}{\mathbb R^n} \newcommand{\dd}{\,\mathrm{d}}$ Consider a pseudodifferential operator (Mellin operator) in positive orthant with symbol $\sigma(z)$: $$ F = \Mel\nolimits^{-1} \sigma \Mel $$ where $\Mel$ is the Mellin transform: $$ f^\ast(z) \equiv (\Mel f)(z) = \int\limits_{\Rn_+}x^{z-I}f(x) \dd x, $$ so that $$ (Ff)(x) = \frac{1}{(2\pi i)^n} \int\limits_{c + i\Rn} x^{-z}\sigma(z) f^\ast(z)\dd z. $$ Let $\widetilde F$ be an other operator which acts on functions $f$ by the following rule: $$ (\widetilde F f)(x) = \frac{1}{(2\pi i)^n} \int\limits_{c + i\Rn}x^{-z} \sigma(I-z)f^\ast(I-z)\dd z, $$ or, introducing operator $(Sg)(z) = g(I-z)$ we can write $\widetilde F = \Mel^{-1} S\sigma \Mel$. Is it true that $\widetilde F$ also has a representation $\widetilde F = \Mel^{-1} \widetilde \sigma \Mel$ with some symbol $\widetilde\sigma$ and if it is true what is the way to find $\widetilde \sigma$?

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