If $g: \mathbb{C} \rightarrow \mathbb{C} $ with $\Delta(g)=\Delta(f)=2 \pi u$ and $\lim_{z \rightarrow \infty} f(z)-g(z)=0.$ Then $g=f.$

Question

Let $u: \mathbb{C} \rightarrow \mathbb{C}$ be a smooth function with compact support. Let $f=2 \pi \log * u $ where $*$ denotes the convolution product. (i.e $\int \log \vert y \vert u(x-y)d \mu_{y}.$) If $g: \mathbb{C} \rightarrow \mathbb{C}$ is a smooth function on $\mathbb{C}$ with $\Delta(g)=\Delta(f)=2 \pi u$ and $\lim_{z \rightarrow \infty} f(z)-g(z)=0.$ Then $g=f.$