For a reaction diffusion equation $u_t=u_{xx}+f(u)$, assume that $\mu_1<\mu_2$ are equilibria and $V(\xi), \xi=x-ct$ with $c>0$ is a traveling front, where $\lim_{\xi\to -\infty}V(\xi)=\mu_2$ and $\lim_{\xi\to\infty}V(\xi)=\mu_1$.
In particular, $\partial_x V(\xi)=V'(\xi)<0$.
Is it possible to say that $$ \lVert V'(\xi-q)\rVert=\sqrt{\langle V'(\xi-q),V'(\xi-q)}=\left(\int_\mathbb{R}e^{c\xi}V'(\xi-q)^2\, d\xi\right)^{1/2} $$
and, similarly, $\lVert V''(\xi-q)\rVert$, are bounded?
(here $q$ is just a fixed real which describes a translation of the front)
The inner product is defined by $$ \langle f,g\rangle=\int_\mathbb{R}e^{c\xi}fg\, d\xi $$ with corresponding norm $$ \lVert x\rVert=\sqrt{\langle x,x\rangle}. $$
Isn't this simply true since each norm is a non-negative real number?
In the case where $\mu_1 = 0, \, \mu_2 = 1, f'(0) = 1$ and $f$ is concave down, the travelling front has the asymptotic behavior $V(\xi) \sim e^{-r\xi}$ with $r = \frac{c - \sqrt{c^2-4}}{2}$. This follows from a linearization argument.
Since $f(V)$ has the same asymptotic behavior as $V$ at $+\infty$, it follows that $V'$ and $V''$ have the same asymptotic behavior at $+ \infty$.
Therefore $\xi \mapsto e^{c\xi} (V')^2(\xi)$ and $\xi \mapsto e^{c\xi} (V'')^2(\xi)$ are not integrable at $+ \infty$.