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


  • $\begingroup$ What is the norm on $H^{1,2}$ again? $\endgroup$ – Julien Feb 15 '13 at 13:15
  • 1
    $\begingroup$ The problem here is, that the operator is not linear so that continuity is not equivalent to boundedness. $\endgroup$ – math12 Feb 15 '13 at 13:53
  • $\begingroup$ You're right. So maybe you should not call this an operator. And also, your claim is not true: continuity is not equivalent to $|Vu|\leq C\|u\|$. You have to bound $Vu_1-Vu_2$, which is not $V(u_1-u_2)$. But what is the $H^{1,2}$ norm? $\endgroup$ – Julien Feb 15 '13 at 13:56
  • $\begingroup$ The $H^{1,2}$-norm is $(\langle u,u\rangle_{L^2}+\langle u',u'\rangle_{L^2})^{1/2}$. $\endgroup$ – math12 Feb 15 '13 at 13:59
  • $\begingroup$ I took for granted that if $u$ is in $H^{1,2}$, then $u^4$ is integrable so that your funtion is well-defined. But how does it follow from the definition of $H^{1,2}$? I am not sure I have your definition of $H^{1,2}$. $\endgroup$ – Julien Feb 15 '13 at 15:33

Let $u,u_0$ be two $H^{1,2}$ functions.


$$ |Vu-Vu_0|\leq \frac{1}{2}\int|(\partial_xu)^2-(\partial_xu_o)^2|dx+\frac{|\alpha|}{2}\int|u^2-u_0^2|dx+\frac{1}{4}\int|u^4-u_0^4|dx. $$

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.