I am reading a paper, which states
Let $\Omega=(x_0,y_0)$ be an open interval in $\mathbb{R}$ and let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a Borel function. Consider the functional
$$ F_\Omega: H^1(\Omega)\longrightarrow [0,+\infty] $$
$$ F_\Omega (u) = \int_\Omega|u'(x)-f(u(x))|^2dx$$If the function $f$ is bounded, then the functional $F_\Omega + \chi_{s_0}$ is coercive, where
$$\chi_{s_0}: H^1(\Omega)\longrightarrow\mathbb{R}$$
is defined by
$$ \chi_{s_0}(u) = \begin{array}{}\left.
+\infty\ \text{ if } u(x_0)\neq s_0 \\
0\ \text{ if } u(x_0)=s_0 \right. \end{array}
$$
I cannot find any argument why the functional is coercive ($\lim_{||u||\rightarrow+\infty}F_\Omega (u)= +\infty)$. Any help will be much appreciated.
Best Answer
First notice that $(F_{\Omega} + \chi_a)(u) \geq F_{\Omega}(u)$ and so it will be enough to show that $F_{\Omega}$ is coercive.
By Poincare's inequality, the norm $\|u\| = \big(\int_\Omega |u'(s)|^2 ds \big)^{\frac12}$ is equivalent to the usual $H^1$-norm and so it is enough to show that $F_\Omega$ is coercive for this norm. For this, we have \begin{align} |F_{\Omega}(u)| = \int_\Omega |u' - f(u)|^2 ds = \int_\Omega |u'|^2 + |f(u)|^2 - 2|f(u) u'| ds \end{align} Since $f$ is bounded, say $|f(x)| \leq C$ for all $x$, we can bound \begin{align} |F_\Omega(u)| \geq \int_\Omega |u'|^2 - 2C |u'| ds \geq \|u\|^2 - 2C |\Omega|^{\frac12} \|u\| \to \infty \end{align} as $\|u\| \to \infty$, as desired (where the last inequality is just an application of Cauchy-Schwartz).