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

Question

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.

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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?

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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?

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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.

Answer 1

Here is one approach:

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 $U(a_{\pm})=0$.