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Given a function space $V$ of some subset of real-valued functions on the real line, linear operator $L: V \rightarrow V$, and $f,g \in V$, define $$ h(t) = \int_{\mathbb{R}}f(u)g(u-t)du $$

Further, assume $h \in V$. Is the below true? $$L(h(t)) = \int_{\mathbb{R}}f(u)L(g(u-t))du $$ If not, under what assumptions is this true? If yes, why?

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See my answer it might be useful for you. – Mhenni Benghorbal Jun 4 '13 at 6:03

1 Answer 1

up vote 1 down vote accepted

An arbitrary operator cannot be moved into the convolution. For example, if $Lh=\psi h$ for some nonconstant function $\psi$, then $$\psi(t) \int_{\mathbb R} f(u) g(u-t) \,du \ne \int_{\mathbb R} f(u) g(u-t) \psi(u-t) \,du $$ for general $f,g$.

However, the identity is true for translation-invariant operators, i.e., those for which $L(g(t-c))=L(g)(t-c)$ for every $c\in\mathbb R$. Indeed, for such operators $$f*(Lg)= \int_{\mathbb{R}}f(u)L(g)(u-t)\,du =\int_{\mathbb{R}}f(u)L(g(u-t))\,du = L(f*g)$$

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