Let $f : \mathbb R^2 \rightarrow \mathbb R $ be a harmonic function. Suppose $$\lim_{\vert x\vert\rightarrow\infty}\frac{\vert f(x)\vert}{\ln\vert x\vert}=0$$ Prove or disprove that $f $ is a constant.
I cannot come up with any example that $f$ is non-constant, so I tend to believe $f$ is constant. From the expression it looks like it is going to use the Liouville Theorem, but I don’t know how to use the $\ln(x)$ condition ($\ln(x)$ is a harmonic function too, but will this be useful?). Are there any ways of dealing this problem? Moreover, I know a problem that if $f$ is continuous and subharmonic with $$\lim_{\vert x\vert\rightarrow\infty}\sup\frac{\vert f(x)\vert}{\ln\vert x\vert}\le0$$ we can also conclude that $f$ is constant. Can we extend the solution of the former problem to the latter problem? Thanks!
The assumption can be weakened to $$\limsup_{|z|\to \infty}{f(x,y)\over \ln|z|}=:r_0<\infty, \ z=x+iy\qquad(*)$$ Fix $r>r_0.$ By $(*)$ we obtain $$f(x,y)-r\ln|z|< 0,\qquad |z|> R$$ for some $R>0.$
Let $f(x,y)=\Re F(x+iy),$ where $F$ is holomorphic. Then $G(z)=\exp \,F(z)$ satisfies $$|G(z)|=\exp(f(x,y))\neq 0, \qquad z=x+iy$$ We have $$ {|G(z)|\over |z|^r}= \exp (f(x,y)-r\ln |z|)<1 ,\qquad |z|>R$$ Thus $$|G(z)|\le |z|^{r},\qquad |z|>R$$ As $|G(z)|$ is bounded for $|z|\le R$ we obtain $$|G(z)|\le |z|^{r}+C_r$$ for a constant $C_r>0.$ Therefore the function $G(z)$ is a polynomial. But $G(z)$ does not vanish, hence $G(z)$ is constant. Therefore $f(x,y)$ is constant.