Let $u: \mathbb{R}^2 \to \mathbb{R}.$ Suppose I have a solution to the equation
$$\Delta(u)+e^{-u} \geq 0$$ on $\mathbb{R}^2$. Let r be the radial coordinate on $\mathbb{R}^2$. Suppose that $$lim_{r \to \infty} u \to c$$
where $c$ is some constant. Can I get some estimate on the maximum of $u$ over all of $\mathbb{R}^2$ ? I'm sorry if my question admits an easy solution, but I am not usually working on analysis and couldn't find this precise situation in the literature.
One cannot bound $\sup u$ in terms of $c$. To see this, set $u(x) = c + \alpha \,\varphi(\varepsilon x)$, where $\varphi\in S(\mathbb R^2;\mathbb R)$, $\varphi(0)=1$, and $\varphi\leq1$, while $\alpha>0$, $\varepsilon>0$ are parameters. Notice that $$ \Delta u(x) + e^{-u(x)} \geq e^{-\alpha} \left( e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c} \right). $$
Now, given $K>c_+$, let $\alpha=K-c>0$ and choose $\varepsilon>0$ in such a way that $e^\alpha\alpha\varepsilon^2 \inf\Delta\varphi + e^{-c}\geq0$. Then $\sup u= u(0)=K$ and $\Delta u + e^{-u}\geq0$; so we have the desired example.
According to J. Liouville Sur l'équation aux différences partielles $\partial^2\log\lambda/\partial u\partial v\pm\lambda/2a^2$ J. Maths. Pures & Appl. 18 (1853), pp 71-71, the general solution of this equation in the plane is given by the formula $$u=-\log\frac{8|f'(z)|^2}{(1-|f(z)|^2)^2}\,,$$where $f$ is holomorphic. You should start from that.
If a solution $u$ exists in the whole plane, it seems that $f$ must be a bounded entire function, hence a constant, from which it follows $u\equiv0$. Perhaps this can be proved by Pohozaev's calculus.
You can also use the paper by H. Fujita Bull. AMS 75 (1969), pp 132-135, which treats the equation in arbitrary space dimension.
