# Must a field approach one of its vacua to have finite energy?

I'm reading these Cornell lectures on solitons (link doesn't work right now, but it just worked yesterday), and I can't seem to prove what I thought would be a simple analysis exercise.

Namely, suppose you have the Lagrangian density

$$\mathcal{L}= \frac{1}{2}(\partial_\mu \phi)^2 - U(\phi)$$

in 1+1 spacetime dimensions with $$g_{\mu\nu} = (1,-1)$$, and where $$\phi$$ is a real scalar field. In the highlighted text attached, the author claims that (i'm paraphrasing)

In order for the integral $$E[\phi] = \int_{-\infty}^\infty \frac{1}{2}\phi^2 + \frac{1}{2}\left(\frac{\partial \phi}{\partial x}\right)^2 + U(\phi)$$ to be finite, then $$U(\phi)$$ must approach a minimum $$\phi_i$$ such that $$U(\phi_i)=0$$ as $$x\to \pm \infty$$.

Question: Is this true? If so why?

My attempt: If the minimum is at $$\phi_i=0$$, then the proof is trivial. So suppose that $$\phi_i \neq 0$$. For simplicity consider only the case that $$\lim_{x\to\infty}\phi(x) = \phi_i$$. Then, for large values of $$x$$ we have that

\begin{align} \frac{1}{2}\phi^2 + \frac{1}{2}\left(\frac{\partial \phi}{\partial x}\right)^2 + U(\phi) &\sim \frac{1}{2}\phi_i^2 + U(\phi_i) \qquad \qquad (\text{large x})\\ &=\frac{1}{2}\phi_i^2 \end{align}

It follows then that the integrand converges to a nonzero value asymptotically, and since the integrand is positive definite we have that $$E[\phi]\to \infty$$ as $$x\to \infty$$. This is in contradiction with the image attached.

Question 2: Is this just a manifestation of the "total energy" divergence that plagues all field theories and really we should be looking at energy differences?

Comments: I am posting this on Pysics SE and not Math SE, because I think the actual answer to this question is an underlying assumption in field theories that I may be missing, not so much a mathematical error.

1. First of all, there is a dot missing on the kinetic term $$\frac{1}{2}\dot{\phi}^2$$. (The typo becomes clear when we compare with the Lagrangian density $${\cal L}$$.) Since we are interested in static configurations, the kinetic term drops out.
2. We will assume that the limits $$\lim_{x\to\infty}\phi(x)=a_+$$ and $$\lim_{x\to-\infty}\phi(x)=a_-$$ exist.
3. We will assume that the potential energy density $$U(\phi)\geq 0$$ is a non-negative continuous function.
4. We will assume that the potential energy functional $$\int_{\mathbb{R}}\mathrm{d}x~U(\phi(x))<\infty$$ is finite.
It is not hard to see that this implies the sought-for conclusion $$U(a_{\pm})=0$$.