Convolution between 2 functions $f$ and $g$ is defined as $$(f\star g)(x) = \int_{-\pi}^\pi f(x') g(x-x') dx'.$$
Shift invariance of convolution is said to be the property that $$f(x-x_0) \star g(x) = (f \star g)(x-x_0).$$
Firstly, what does $f(x-x_0) \star g(x)$ represent and why $ f(x-x_0) \star g(x) = (f \star g)(x-x_0)$?
$$(f\star g)(x-x_0) = \int_{-\pi}^\pi f(x') g(x-x_0-x') dx'.$$ Aplly the transformation of variables $y=x'+x_0$, then $dy=dx'$ and the integral becomes $$\int_{-\pi+x_0}^{\pi+x_0} f(y-x_0) g(x-y) dy$$ But $$f(x-x_0) \star g(x)=\int_{-\pi}^{\pi} f(y-x_0) g(x-y) dy$$ Then if $f$ is periodic with period $2\pi$ then $f(x-x_0) \star g(x) = (f \star g)(x-x_0)$