Let $\Omega \subset \mathbb{R}^2$ and consider the following operator: $a : H^1(\Omega) \to H^1(\Omega)$ $$a(u, v) = \int_\Omega \nabla u \nabla v \; dV + \int_{\partial \Omega} \frac{\alpha}{\beta} u v \; d\sigma.$$ $\alpha, \; \beta : \partial \Omega \to \mathbb{R}$, $\alpha \beta \neq 0$, $\frac{\alpha}{\beta} > 0, \; \frac{\alpha}{\beta}, \frac{1}{\beta} \in L^{\infty} (\Omega), \; \frac{\alpha (x,y)}{\beta(x, y)} \geq c_0 >0.$
I have to show that $a$ is continuous, i.e $\exists \; c>0$ such that $|a(u,v)| \leq c ||u||_{H^1 }||v||_{H^1}, \; \forall \; u, v \in H^1(\Omega)$.
My work:
$$|a(u, v)| \leq \int_{\Omega}|\nabla u \nabla v| dV + \int_{\partial \Omega} |\frac{\alpha}{\beta}| |u v| d\sigma.$$ But $$\nabla u \nabla v = \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial v}{\partial y}$$ and $$\frac{\alpha}{\beta} \in L^\infty (\Omega) \Rightarrow |\frac{\alpha}{\beta}| \leq ||\frac{\alpha}{\beta}||_{L^\infty} =: a.$$ Then $$|a(u,v)| \leq \int_{\Omega}|\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}| + \int_{\Omega}|\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}| + a \int_{\partial \Omega} |uv| d\sigma \underset{Holder}{\leq}$$ $$ \leq ||\frac{\partial u}{\partial x}||_2 ||\frac{\partial v}{\partial x}||_2 + ||\frac{\partial u}{\partial y}||_2 ||\frac{\partial v}{\partial y}||_2 + a ||u||_2 ||v||_2.$$ $$||u||_{H^1} ||v||_{H^1} = (||u||^2_2 + ||\frac{\partial u}{\partial x}||_2^2 + ||\frac{\partial u}{\partial y}||_2^2)^{\frac{1}{2}}(||v||^2_2 + ||\frac{\partial v}{\partial x}||_2^2 + ||\frac{\partial v}{\partial y}||_2^2)^{\frac{1}{2}} \geq (||\frac{\partial u}{\partial x}||_2^2 ||\frac{\partial v}{\partial x}||_2^2 + ||\frac{\partial u}{\partial y}||_2^2 ||\frac{\partial v}{\partial y}||_2^2 + ||u||_2^2 ||v||^2_2)^\frac{1}{2}.$$
I don't know what to do next. Can someone help me?
Thank you!
