Constructing a locally integrable function

Question

Let $\epsilon\in(0,1)$ and $F^{\epsilon}:\mathbb{R}^2\to\mathbb{R}$ defined by $$F^{\epsilon}(x)=\log(|x|^2+\epsilon^2)$$ How can I construct a $g \in L^1_{loc}({\mathbb{R}}^2)$ such that $$|F^{\epsilon}(x)|\leq g(x), \ \ \forall x\in \mathbb{R}^2, \ \ \epsilon\in(0,1). $$

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If $||x||>1$, it's easy to see that $g(x)=\log(|x|^2+1)$ do the job. My problem is near the origin.

Answer 1

What about $g(x) =|\log(|x|^2)|+\log(2),$ for $|x|\leq 1$?

Clearly $g(x)\in L^1_{loc}({\mathbb R}^2)$ (use polar coordinates to check) and $$|F^{\epsilon}(x)|\leq g(x),\forall x{\rm ~with~}|x|\leq 1,\epsilon\in (0,1).$$