Let $\Omega$ a bounded connected open regular set, and let $f \in L^2(\Omega)$. We consider the variational problem: find $u \in H^1(\Omega)$ such: $$\displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\displaystyle\int_{\Omega} v dx\right) = \displaystyle\int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$ with: $$\exists \alpha > 0, A(x) \xi \xi \geq \alpha |\xi|^2, \forall \xi \in \mathbb{R}^n, \exists \beta > 0, |A(x) \xi| \leq \beta |\xi|, \forall \xi \in \mathbb{R}^n$$

The question is: prove that this variational problem admits a unique solution $u \in H^1(\Omega)$.

My solution is: we put $$a(u,v) = \displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\int_{\Omega} v dx\right)$$ $$L(v) = \displaystyle\int_{\Omega} f v dx.$$

To prove continuity of $a$, we have to prove that $$\exists M > 0, |a(u,v)| \leq ||u||_{H^1(\Omega)} ||v||_{H^1(\Omega}$$ Let $u \in H^1(\Omega)$, and let $v \in H^1(\Omega)$. $|a(u,v)| = |\displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + (\displaystyle\int_{\Omega} u dx)(\displaystyle\int_{\Omega} v dx)|$

$|a(u,v)| \leq \displaystyle\int_{\Omega} |A \nabla u \cdot \nabla v| dx + (\displaystyle\int_{\Omega} |u|dx) (\displaystyle\int_{\Omega} |v|dx)$

we have by Cauchy-Schwarz inequality that: $\displaystyle\int_{\Omega}|A \nabla u \cdot \nabla v| dx \leq (\displaystyle\int_{\Omega} |A \nabla u|^2 dx)^{1/2} (\displaystyle\int_{\Omega} |\nabla v|^2 dx)^{1/2}$ $\leq \beta (\displaystyle\int_{\Omega} |\nabla u|^2 dx)^{1/2} (\displaystyle\int_{\Omega} |\nabla v|^2 dx)^{1/2} \leq \beta ||\nabla u||_{L^2} ||\nabla v||_{L^2}$

so $$\displaystyle\int_{\Omega} |A \nabla u \cdot \nabla v| dx \leq \beta ||u||_{H^1} ||v||_{H^1}$$

and we have by Cauchy-Schwarz inequality: $(\displaystyle\int_{\Omega} |u|dx)(\displaystyle\int_{\Omega} |v|dx) \leq ||u||_{L^2} ||v||_{L^2} \leq ||u||_{H^1} ||v||_{H^1}$ then, we conclude that $|a(u,v)| \leq \beta ||u||_{H^1} ||v||_{H^1} + ||u||_{H^1} ||v||_{H^1} = (\beta + 1) ||u||_{H^1} ||v||_{H^1}$ then, there exists $M = \beta + 1$ such $|a(u,v)| \leq M ||u||_{H^1} ||v||_{H^1}$.

How can I write a perfect proof of continuity of $a$?

Who can help me please, in my last question.

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    $\begingroup$ Why registering a new account? math.stackexchange.com/questions/434702/… $\endgroup$ – Shuhao Cao Jul 3 '13 at 19:33
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    $\begingroup$ And your proof of continuity is ok, but you might wanna add a factor of $|\Omega|$ in front of the estimate for $(\int_{\Omega} u)(\int_{\Omega} v)$, and absolute value is not necessary. $\endgroup$ – Shuhao Cao Jul 3 '13 at 19:35
  • $\begingroup$ beacause i can't open my old acount $\endgroup$ – jijii Jul 3 '13 at 21:47
  • $\begingroup$ how i can add a factor $|\Omega|$? and thoma tell me in a precedent topic that my proof is not complete beacause there are some constants missing and some details you have to improve. What's an perfect and complete proof please. $\endgroup$ – jijii Jul 3 '13 at 21:49
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    $\begingroup$ You can always use "forgot password" option. Also when you use the Cauchy-Schwarz: $$\int_{\Omega} v\,dx \leq \left(\int_{\Omega} v^2\,dx\right)^{1/2} \left(\int_{\Omega} 1\,dx\right)^{1/2},$$ and the second term is the square root of the measure (area, volume,etc) of $\Omega$, together with the integral term for $u$ you have that factor. $\endgroup$ – Shuhao Cao Jul 3 '13 at 22:19

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