Pulling Operator Inside Integral

Question

Say $Y$ is a Banach space and you have a family of continuous/bounded operators $L_{x}: Y \rightarrow Y$ for $x\in \mathbb{R}$ and say you have an bounded, smooth map $f(x):\mathbb{R}\rightarrow Y$. How can I show, as we do for derivative operators, that $L_{x}\int_{\mathbb{R}}{L_{y}f(y)dy} = \int_{\mathbb{R}}{L_{x}L_{y}f(y)dy}$.

Answer 1

It depends on how you define $Y$ valued integrals.

Such integrals usually satisfy (or are defined by) the condition that $\phi(\int g(x) d \mu (x)) = \int \phi(g(x)) d \mu (x)$ for all $\phi \in Y^*$.

If this is the case, then we have \begin{eqnarray} \phi(\int_{\mathbb{R}}{L_{x}L_{y}f(y)dy} ) &=& \int_{\mathbb{R}}{\phi(L_{x}L_{y}f(y))dy} \\ &=& \int_{\mathbb{R}}{(\phi \circ L_{x})(L_{y}f(y))dy} \\ &=& (\phi \circ L_x) (\int_{\mathbb{R}}{L_{y}f(y)dy} ) \\ &=& \phi(L_x (\int_{\mathbb{R}}{L_{y}f(y)dy} )) \end{eqnarray} where we have used the fact that $\phi \circ L_x \in Y^*$.

In other words, $\int_{\mathbb{R}}{L_{x}L_{y}f(y)dy} = L_x (\int_{\mathbb{R}}{L_{y}f(y)dy} )$.