Suppose $f,g\in L^1(\mathbb{R})$ are non negative, even and decreasing on $[0,\infty)$. Prove that their convolution $f * g$ is even and decreasing on $[0,\infty)$.
Attempt: It is easy to show that $f * g$ is even, this simply follows from a change of variables. But how to show that it is decreasing? Let $0\leq x<z$. Then:
$f*g(x)=\int_{\mathbb{R}} f(y)g(x-y)dy$
$f*g(z)=\int_{\mathbb{R}} f(y)g(z-y)dy$
I need somehow to show that the first integral is at least as large as the second one. I tried to split the domain into parts. Since $f$ is non negative multiplication by it preserves inequalities. So if $y<x$ then $f(y)g(x-y)\geq f(y)g(z-y)$. If $y>z$ then $f(y)g(x-y)\leq f(y)g(z-y)$ (since on the negative real line the function $g$ is increasing, follows from it being even). So taking these inequalities inside the integrals gives us:
$\int_{y<x}f(y)g(x-y)\geq\int_{y<x}f(y)g(z-y)dy$
And there is a reverse inequality on $\{y>z\}$. So intuitively the interval $(-\infty, x)$ is "longer" than $(z,\infty)$ and contains more values near the origin, which is where $f$ and $g$ are large. However, I don't see how can it be written formally. Also, there remains the interval $\{x\leq y\leq z\}$ where there are no such inequalities, so it's difficult to say what does it add to both integrals. Any ideas?