If $k$ is an squared or absolutely integrable kernel are the belo equalities true ? $$z(s)=\int_{R}^{} \! k(u-d) x(u).du \ \ =\int_{R}^{} \! k(u+d) x(u).du \ \ $$
and
$$\int_{R}^{} \! k(u-d) k(u-d^{'}).du \ \ =\int_{R}^{} \! k(d-u) k(d^{'}-u).du \ \ $$
Consider $k(u) = \delta(u)$ the delta function. If you do not like this, you can find a sequence of smooth rapidly decaying functions that approximate it. Then, $\int k(u-d) x(u) du = x(d)$ while $\int k(u+d) x(u) du = x(-d)$, assuming $x$ is integrable.
The other one will hold if the kernel is symmetric.
