Why can kink not tunnel to the vacuum, making it topologically stable? Why can the kink
$$\phi(x)=v\tanh\left(\frac{x}{\xi}\right)$$
not tunnel into vacuum $+v$ or $-v$ (with spontaneous symmetry breaking in the vacuum)?
From the boundary condition, $\phi(x)\rightarrow \pm v$ as $x\rightarrow\pm\infty$, it is self-evident.  However, the book states:

Due to the infinite high energy barrier, the kink cannot tunnel into the vacuum.

Where is the infinite high energy barrier? The energy density is $$E(x)=\frac{gv^4}{2}{\rm sech}^4\left(\frac{x}{\xi}\right),$$
whose integration over all space is finite.
Where is the infinite high energy barrier?
 A: Here we assume that OP's question asks about $\phi^4$-theory in 1+1D, where the lagrangian density reads
$$\tag{1} {\cal L}~=~\frac{1}{2}\dot\phi^2 -{\cal U}, \qquad 
{\cal U}~:=~ \frac{1}{2} \phi^{\prime 2} + {\cal V},\qquad  
\phi \in C^1(\mathbb{R}^2),$$ 
where the $\phi^4$-potential density
$$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ 
has two minimum points at $\phi=\pm v$, i.e. a double-well. In eq. (1) the dot (prime) means differentiation wrt. $t$ ($x$), respectively. 

We rephrase OP's question as follows:

Prove that there don't exist finite energy homotopies $\phi:\mathbb{R}^2\to \mathbb{R}$ between the following 4 topological sectors: the kink, the antikink, and the two vacuum solutions $\phi=\pm v$. 

Here the kink has limits $$\tag{3} \lim_{x\to \pm \infty}\phi(x)~=~\pm v,$$ and the antikink has limits $$\tag{4} \lim_{x\to \pm \infty}\phi(x)~=~\mp v. $$
Sketched indirect proof: Assume that a homotopy $\phi$ exists. To be concrete, say, between the kink and the left vacuum solution $\phi=-v$. So the homotopy $\phi$ has to change valley for positive $x$. Since this is Phys.SE rather than Math.SE, we are for simplicity going to assume that for arbitrary instants $t\in\mathbb{R}$, the limits
$$\tag{5} f_+(t)~:=~\lim_{x\to \infty}\phi(x,t) \quad\text{and}\quad f_-(t)~:=~\lim_{x\to -\infty}\phi(x,t) ,  $$ 
exist. Then to have finite potential energy 
$$\tag{6} V(t)~:=~ \int_{\mathbb{R}}\! dx~{\cal V}(\phi(x,t))~<~\infty,$$
it follows that the two functions $f_{\pm}(t)$ can only take the values $\pm v$. Intuitively $\phi$ is then topologically confined to the two potential valleys for sufficiently large $x$.  It follows that there exists a sufficiently large constant $K$ such that $\forall x\geq K $ the function $t \mapsto \phi(x,t)$ cannot be continuous in $t$. Contradiction.
References:


*

*S. Coleman, Aspects of Symmetry, 1985; Section 6.3.1.

*R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987; Sections 2.3-2.4.
A: The energy density of the state $\pm v$ is going to be something like $\propto μ^4$, if you are using the basic $\varphi^4$ theory. While the energy of the domain wall is finite, the energy of the vacuum state is not, and so the transition to the vacuum state iver all space will be infinite.
