I'm having one big problem solving one exercise our professor gave us. I have an $u\in\mathcal{C}(\mathbb{R^{n+1}})$ such as $$\liminf_{|z|\to\infty}u(z)\geq 0$$ and I have to show that $u\geq 0$. With some help from friends and notes from my professor I found the solution of the exercise but there's a passage that's keeps being obscure to me. Can I show that if $\liminf_{|z|\to\infty} u(z)\geq0$ then $u$ has an $\inf$.
I've spent the last three days asking to some firend of mine but we didn't come to a conclusion. Can anyone help me?.
Ps.:
We call $u\in\mathcal{C}(\Omega)$ as a caloric function and it's defined as follow:
$u\in\mathcal{C}(\Omega)$ iff $u:\Omega\to\mathbb{R}$, with $\Omega\subseteq \mathbb{R}^{n+1}$ it's an open subset, and $u$ has the two following properties:
- $u\in C^{2,1}(\Omega,\mathbb{R})$ i.e.: $\partial_{x_j}u,\;\partial_{x_jx_i}u,\;\partial_{t}u$ exisist in any point of $\Omega$ and are continuos function in $\Omega$, for all $i,j=\lbrace1,2,...,n\rbrace$
- $Hu(z)=\triangle u(z)-\partial_t u(z)=0\;\;\;\;\forall z\in \Omega$
Because $\liminf_{|z|\to \infty} u(z) \ge 0$, then for any $\epsilon > 0$, there is an $R$ such that if $|z| > R$, then $u(z) > -\epsilon$. For this, it is sufficient to have a fixed value, say $\epsilon = 1$. So for $|z| > R, u(z) > -1$.
But $u$ is continuous on the closed ball $\overline B_0(R) = \{z \mid |z| \le R\}$. That ball is closed and bounded, and therefore compact. The continuous image of a compact set is itself compact, so $u(\overline B_0(R))$ is a compact subset of $\Bbb R$. Any such set is bounded, so $u$ has some lower bound on $\overline B_0(R)$ and a lower bound of $-1$ on its complement. Therefore $u(\Bbb R^{n+1})$ is bounded below and must have an infimum.