Upper bounding convolution between compactly supported function and logarithm

Question

Let $h:\mathbb{R}^2\rightarrow \mathbb{C}$ be a $C^\infty$ function with compact support which satisfies $\int h(z)\mathrm{d}x\, \mathrm{d}y = 0$. How can one show that $$f(w) = \log \frac{1}{|\cdot|}\ast h(w) = \int_{\mathbb{R}^2}\log \frac{1}{|z|}h(w-z)\mathrm{d}x\, \mathrm{d}y$$ satisfies $|f(w)|\leq \frac{C}{|w|}$ with some bound $C$.

Answer 1

I came up with my own solution:

We have that

$$f(w) = \int_{\mathbb{R}^2}\log \frac{1}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y = \int_{\mathbb{R}^2}\log \frac{|w|}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y $$

by the specified condition. Now we may assume that $h$ has support in $B(0,R)$ and therefore for large enough $w$

$$|f(w)w|\leq \int_{B(0,R)}|w|\log \frac{|w|}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y\leq \int_{B(0,R)}|w|\log \frac{|w|}{|w|-R}h(z)\mathrm{d}x\, \mathrm{d}y = |w|\log \frac{|w|}{|w|-R}\int_{B(0,R)}h(z)\mathrm{d}x\, \mathrm{d}y.$$

The term $x\log \frac{x}{x-R}$ has the finite limit $R$ as $x\rightarrow \infty$.