Let $R$ be the ring of distributions $T\in \mathcal{D}'(\mathbb{R})$ with support in $[0,\infty)$ and with the operations of pointwise addition and multiplication taken as convolution, and $I$ be the ideal in $R$ with support in $(0,\infty)$. Is $I$ maximal in $R$?
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3$\begingroup$ Is convolution really well-defined in this setting? (double checking, it seems so) $\endgroup$– YCorCommented Jun 14, 2018 at 9:03
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1$\begingroup$ How do you define the support of a distribution? For me this is a closed subset of $\Bbb{R}$. $\endgroup$– abxCommented Jun 14, 2018 at 9:27
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$\begingroup$ Of course, the support of a distribution is closed. If it is contained in $(0,\infty)$ there is a strictly positive $a$ such that it is contained in $[a,\infty)$. $\endgroup$– Jochen WengenrothCommented Jun 14, 2018 at 14:22
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2 Answers
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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 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 $v\in I$.
answered Jun 14, 2018 at 10:28
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Consider the ideal $J$ generated by all continuous locally integrable functions $f:[0,\infty)\to\mathbb{R}$ with $f(0) = 0$ (considered as distributions). Then the ideal $I+J$ is different from $R$, as it does not contain the delta distribution at 0. On the other hand, $I + J$ strictly contains $I$.
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$\begingroup$ I doubt that $J$ is an ideal! The function $f(x)=x$ for $x\ge 0$ and $f(x)=0$ for $x\le 0$ belongs to $J$ and $\delta_0'' \ast f = f'' = \delta_0$. $\endgroup$ Commented Jun 14, 2018 at 10:22
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$\begingroup$ Well, $J$ is an ideal, by definition. But maybe it is not a strict ideal. Ok, I was wrong. $\endgroup$– jarauhCommented Jun 14, 2018 at 10:26
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$\begingroup$ @jarauh By definition, the full ring isn't an ideal. Otherwise it would always and trivially be the maximal ideal. $\endgroup$ Commented Jun 14, 2018 at 10:32
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3$\begingroup$ @JochenWengenroth I disagree. The full ring is an ideal. Otherwise, the sum of ideals would not be an ideal, etc., and the definition of "the ideal generated by..." would be difficult. On the other hand, the definition of a maximal ideal explicitly excludes the full ring. That is, "maximal ideal" is short hand for "maximal proper ideal". $\endgroup$– jarauhCommented Jun 14, 2018 at 10:36
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