Let us think of the Hilbert space \begin{equation} H:=\Bigl( f \in L^2[0,1] \mid \int_0^1 f(x) dx=0 \Bigr) \end{equation} that is, the space of $L^2$ functions without the zero mode.
On $H$, I am aware that the inverse of the Laplacian $(-\Delta)^{-1}$ can be defined as a bounded linear map. In fact, $(-\Delta)^{-1}$ maps $H$ into the Sobolev space $H^2[0,1]$.
Now, for any $f, g \in H$, let us denote by $f*g$ their convolution: \begin{equation} (f*g)(x)=\int_0^1 f(x-y)g(y)dy \end{equation}
It is clear from Fourier transform that $f*g \in H$ as well.
My question is that, the following manipulation is correct?: \begin{equation} (-\Delta)^{-1}(f*g)=\Bigl[(-\Delta)^{-1}f \Bigr] * g = f* \Bigl[(-\Delta)^{-1} g\Bigr] \end{equation}
I mimicked the property of differential operation on convolution but not sure if this is really true.. Could anyone please help me?