Convolution with the "wrong" sign

Question

Let $u\in L^\infty(\mathbb R^n)$, $\eta\in C([-1, 1]^n, [0, 1])$ and, given $\epsilon>0$ consider the mollifier $\eta_\epsilon(x) = \epsilon^{-n} \eta(x/\epsilon)$.

Define $u_\epsilon = u\ast \eta_\epsilon$, where $\ast$ denote the convolution https://en.wikipedia.org/wiki/Convolution#:~:text=The%20term%20convolution%20refers%20to,the%20y%2Daxis%20and%20shifted.

Let $s\in )0, 1)$. For an exercise, I am calculating $\frac{(u_\epsilon(x)-u_\epsilon(y))}{|x-y|^{n+2s}}$ which according to me is $$\frac{(u_\epsilon(x)-u_\epsilon(y))}{|x-y|^{n+2s}} =\int_{\mathbb R^n} \frac{(u(x-z) +u(y-z)) \eta_\epsilon(z)}{|x-y|^{n+2s}} dz.$$

Reading the solution proposed by the lecturer, the idea to solve the exercise is the same as mine, but when he computes the convolution writes $$\frac{(u_\epsilon(x)-u_\epsilon(y))}{|x-y|^{n+2s}} =\int_{\mathbb R^n} \frac{(u(x+z) +u(y+z)) \eta_\epsilon(z)}{|x-y|^{n+2s}} dz.$$

Are the two definitions equivalent? I do not understand why he uses the "+" sign, I have tried with a change of variables and other tricks, but it does not work, unless one assume any symmetry on $\eta_\epsilon$.

Anyone please help me to understand why the latter inequality holds? How to get from $\int_{\mathbb R^n} \frac{(u(x-z) +u(y-z)) \eta_\epsilon(z)}{|x-y|^{n+2s}} dz$ to $\int_{\mathbb R^n} \frac{(u(x+z) +u(y+z)) \eta_\epsilon(z)}{|x-y|^{n+2s}} dz$?

Answer 1

Most of the time, it is fine to assume the mollifier is a radial function, so $\eta(z) = \eta(-z)$. I can't see why it wouldn't be fine to assume so here. Otherwise, like you said, the two integrals generally won't be the same, and I would use the convention you originally made, rather than the one with "$+$" signs.