convolution of nonvanishing function

Question

I guess function $f$, defined by $$f(x)=\int_{\mathbb{R}^2}\log|x-y|\exp(-|y|^2)dy,$$ is 'not' vanishing at infinity. But I don't know how to prove rigorously.

Answer 1

Hint: $\log|y|=s(y)+g(y)$, where $s$ is integrable (with compact support) and $g(y)\ge1$ for all $y$. Now if $h(x)=e^{-|x|^2}$ then the convolution $s*h$ does vanish at infinity and $g*h$ clearly does not.

Answer 2

Let $L_N(x) = \min(N, \log(|x|))$ be a "cutoff" logarithm.
Note that $L_N$ is not bounded below, but the negative part is integrable. Write $L_N(x) = A_N(x) + B(x)$ where $B(x) =\log(|x|)$ where $|x| < 1$, $0$ otherwise, and $A_N(x) = L_N(x)$ where $|x|\ge 1$, $0$ otherwise.

Now $$ f(x) \ge \int_{\mathbb R^2} L_N(|x-y|) \exp(-|y|^2)\; dy$$ As $|x| \to \infty$, by the Dominated Convergence Theorem we have $$ \int_{\mathbb R^2} A_N(|x-y|) \exp(-|y|^2)\; dy \to \int_{\mathbb R^2} N \exp(-|y|^2)\; dy = N \pi $$ while for $|x| > 1$, $$ 0 \ge \int_{\mathbb R^2} B(|x-y|) \exp(-|y|^2)\; dy \ge \exp(-(|x|-1)^2) \int_{|u|<1} \log |u|\; du $$ which goes to $0$ as $|x| \to \infty$. Thus $$\int_{\mathbb R^2} L_N(|x-y|) \exp(-|y|^2)\; dy \to N \pi\ \text{as} |x| \to \infty$$ and this implies that $f(x) \to \infty$ as $|x| \to \infty$.