Show that operator is continuous 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
 A: 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.
A: I am assuming that the norm is given by $$\|u\|_{1,2}=\|u\|_2+\|\nabla u\|_2$$
One way to prove continuity is to show that if $u_n\rightarrow u$ in $H^{1,2}$, then $Vu_n\rightarrow Vu$ in $\mathbb{R}$.
Theorem (see Brezis Theorem 4.9): Let $f_n$ be a sequence in $L^p$ and let $f\in L^p$ be such that $\|f_n-f\|_p\rightarrow 0.$ Then there exist a subsequence $(f_{{n_k}})$ and a function $h\in L^p$ such that :
a) $f_{n_k}\rightarrow f$ a.e.
b) $|f_{n_k}|\leq h$ 
Now, suppose that $u_n\rightarrow u$in $H^{1,2}$. Then, $u_n\rightarrow u$ in $L^2$. By using the theorem, we can suppose that there exist $h\in L^1$ such that $u_n\rightarrow u$ a.e. and $|u_n|^2\leq h$. We can also suppose that $h\in L^\infty$ because the continuous injection of $H^{1,2}$ in $L^\infty$.
It follow that $u_n^4\rightarrow u^4$ a.e.. I'm gonna prove that there exist $g\in L^1$ such that $|u_n|^4\leq g$. Indeed, in the set $\{x: h\leq 1\}$, we have that $h^2\leq h$. On the other hand, because $h\in L^\infty$, we can modify $h$ (to some $\tilde{h}$) in the set $\{x: h\geq 1\}$ in such a way that $u^4\leq \tilde{h}$ in this set. By joining $h$ and $\tilde{h}$ and calling it $g$ we have that $u^4\leq g$, where $g\in L^1$.
Now you can apply Lebesgue theorem to conclude that $Vu_n\rightarrow Vu$ in $\mathbb{R}$.
Note: There is some proposital gaps in the demonstration. Try to solve it and if you have some doubt, please post here.
