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Jochen Wengenroth
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The set $J=\{u\in\mathscr D'(\mathbb R):$ supp$u \subseteq [0,\infty)$ and singsupp$u \subseteq (0,\infty)\}$ is a strictly bigger ideal.

To see that it is aan ideal decompose such a $u$ by multiplying with a cut-off function as $u=\varphi + v$ where $\varphi \in \mathscr D([0,\infty))$ and $v\in I$.

The set $J=\{u\in\mathscr D'(\mathbb R):$ supp$u \subseteq [0,\infty)$ and singsupp$u \subseteq (0,\infty)\}$ is a strictly bigger ideal.

To see that it is a ideal decompose such a $u$ by multiplying with a cut-off function as $u=\varphi + v$ where $\varphi \in \mathscr D([0,\infty))$ and $v\in I$.

The set $J=\{u\in\mathscr D'(\mathbb R):$ supp$u \subseteq [0,\infty)$ and singsupp$u \subseteq (0,\infty)\}$ is a strictly bigger ideal.

To see that it is an ideal decompose such a $u$ by multiplying with a cut-off function as $u=\varphi + v$ where $\varphi \in \mathscr D([0,\infty))$ and $v\in I$.

Source Link
Jochen Wengenroth
  • 16.4k
  • 2
  • 42
  • 82

The set $J=\{u\in\mathscr D'(\mathbb R):$ supp$u \subseteq [0,\infty)$ and singsupp$u \subseteq (0,\infty)\}$ is a strictly bigger ideal.

To see that it is a ideal decompose such a $u$ by multiplying with a cut-off function as $u=\varphi + v$ where $\varphi \in \mathscr D([0,\infty))$ and $v\in I$.