Let $T: C_{0}^{\infty}(R^{n}) \to C^{\infty}(R^{n})$ be a linear operator. $T$ is local if $$\operatorname{supp} (Tu) \subset \operatorname{supp} (u),$$ for all $u \in C_{0}^{\infty}({R}^{n})$. We know all differential operators are local. But what is an example of non local operator?
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For example $$ u \mapsto \int_{\mathbb R^n} u \, dx \cdot 1_{\mathbb R^n} $$ is a linear operator, which is non-local.
martini
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1It's also interesting to note that this fact leads to the fractional derivative being non-local. – Dan Jun 16 '14 at 09:29
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@martini Do you mean characteristic function by your "$1_{R^{n}}$"? – curiousm Jun 16 '14 at 09:33
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Yes, the constant function $1_{\mathbb R^n}\colon x \mapsto 1$. – martini Jun 16 '14 at 09:34
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1Thanks you so much! But why do we need to multiply by $1_{R^{n}}$? – curiousm Jun 16 '14 at 09:42
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@curiousm Just because you want $C^\infty_0 \to C^\infty$, but not $C_0^\infty \to \mathbb R$. – Jun 16 '14 at 10:30