# Under What Conditions and Why can Move Operator under Integral?

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?

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

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 $$f*(Lg)= \int_{\mathbb{R}}f(u)L(g)(u-t)\,du =\int_{\mathbb{R}}f(u)L(g(u-t))\,du = L(f*g)$$