Define a convolution operator which takes two functions $f,g : \mathbb{R}^d \to \mathbb{R}$ and gives a function $f*g: \mathbb{R}^d \to \mathbb{R}$ defined as $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y) \, dy$$ Show that $f*g$ is uniformly continuous when $f$ is integrable and $g$ is bounded.
I have a question about the proof for the uniformly continuous part. The solution was posted at https://analysisqualifier.blogspot.com/2013/06/224.html
The proof goes as follows:
My question is regarding the step highlighted in red. From my understanding it appears to me that they're subtracting out an $x$ at this step. Is this because 1.) $x$ is a constant and 2.) the integral is translation invariant over $\mathbb{R}^d$? I just want to make sure I'm understanding this step correctly.
It's the change of variables $y \mapsto y + x$. Perhaps it becomes clearer if we set $u = y-x$ and observe that $du=dy$ and the limits of integration remain the same and the integral becomes $$\int_{\mathbb{R}^d}|f(-u) - f(z-x - u)| du$$ but since $u$ is simply an integration variable, we could re-label it as $y$ if we wished.