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If $f$ a distribution with compact support then there are they exist$m$ and measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
how to demonstrate this result ?
If $f$ a distribution with compact support then there are$m$ and measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
how to demonstrate this result ?
If $f$ a distribution with compact support then they exist$m$ and measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
If $f$ a distribution with compact support then there are$m$ and are measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
how to demonstrate this result ?
If $f$ a distribution with compact support then there $m$ and are measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
how to demonstrate this result ?
If $f$ a distribution with compact support then there are$m$ and measures $f_\beta$,$|\beta|\leq m$ such that
$$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$
