Show that
$$
V\colon H^{1,2}(\mathbb{R},\mathbb{R})\to\mathbb{R}
$$
is continuous, where
$$
V(u)=\int\limits_{-\infty}^{\infty}\left(\frac{1}{2}(\partial_x u)^2-\frac{\alpha}{2}u^2+\frac{1}{4}u^4\right)(x)\, dx.
$$
To my knowledge, I have to show
$\lvert Vu\rvert\leq\lVert u\rVert_{H^{1,2}}\cdot C$ for a constant $C\geq 0$.
I am not sure if I am right when doing this:
$\lvert Vu\rvert=\left\lvert\int\limits_{-\infty}^{\infty}\left(\frac{1}{2}(\partial_x u)^2-\frac{\alpha}{2}u^2+\frac{1}{4}u^4\right)(x)\, dx\right\rvert$
$\leq\int\limits_{-\infty}^{\infty}\left\lvert\left(\frac{1}{2}(\partial_x u)^2-\frac{\alpha}{2}u^2+\frac{1}{4}u^4\right)(x)\right\rvert\, dx$
$\leq\frac{1}{2}\int\limits_{-\infty}^{\infty}\lvert u'(x)^2\rvert\, dx+\frac{1}{4}\int\limits_{-\infty}^{\infty}\lvert u^4(x)\rvert\, dx+\frac{\lvert\alpha\rvert}{2}\int\limits_{-\infty}^{\infty}\lvert u^2(x)\rvert\, dx$
$\leq\int\limits_{-\infty}^{\infty}\lvert u'(x)^2\rvert\, dx+\int\limits_{-\infty}^{\infty}\lvert u^4(x)\rvert\, dx+\frac{\lvert\alpha\rvert}{2}\int\limits_{-\infty}^{\infty}\lvert u^2(x)\rvert\, dx$
$=\langle u',u'\rangle_{L^2}+\int\limits_{-\infty}^{\infty}\lvert u^4(x)\rvert\, dx+\frac{\lvert\alpha\rvert}{2}\langle u,u\rangle_{L^2}$
If this is right (I do not think so…): What do I have to do now?
With regards
math12
Best Answer
Let $u,u_0$ be two $H^{1,2}$ functions.
Then
$$ |Vu-Vu_0|\leq \frac{1}{2}\int|(\partial_xu)^2-(\partial_xu_o)^2|dx+\frac{|\alpha|}{2}\int|u^2-u_0^2|dx+\frac{1}{4}\int|u^4-u_0^4|dx. $$
For the first term, we have, by Cauchy Schwarz, $$ \int|\partial_x(u-u_0)||\partial_x(u+u_0)|dx\leq \|\partial_x(u-u_0)\|_2\|\partial_x(u+u_0)\|_2\leq\|u'-u_0'\|_2\|u'+u_0'\|_2 $$ so it is bounded by $$ \|u-u_0\|\|u+u_0\| $$ which tends to $0$ as $u$ tends to $u_0$ in $H^{1,2}$.
You can treat the other terms in a similar fashion, essentially by factoring $ u^2-u_0^2=(u-u_0)(u+u_0) $ and $u^4-u_0^4=(u-u_0)(u^3+u^2u_0+uu_0^2+u^3)$, and then using Cauchy Schwarz.