Showing that the following operator is continuous 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!
 A: You have shown that
$$|a(u,v)| \leq \|u_x\|_2 \|v_x\|_2 + \|u_y\|_2 \|v_y\|_2 + a \|u\|_2 \|v\|_2.\tag{1}$$
So, from your last inequality:
\begin{align}
\|u\|_{H^1} \|v\|_{H^1} &= (\|u\|^2_2 + \|u_x\|_2^2 + \|u_y\|_2^2)^{\frac{1}{2}}(\|v\|^2_2 + \|v_x\|_2^2 + \|v_y\|_2^2)^{\frac{1}{2}}\\
& \geq (\|u_x\|_2^2 \|v_x\|_2^2 + \|u_y\|_2^2 \|v_y\|_2^2 + \|u||_2^2 \|v\|^2_2)^\frac{1}{2}\\
&\overset{(*)}{\geq} C (\|u_x\|_2 \|v_x\|_2 + \|u_y\|_2 \|v_y\|_2 + \|u\|_2 \|v\|_2)\\
&\geq C \frac{1}{\min\{1,a\}}(\|u_x\|_2 \|v_x\|_2 + \|u_y\|_2 \|v_y\|_2 + a\|u\|_2\|v\|_2) \\
&\overset{(1)}{\geq} C \frac{1}{\min\{1,a\}} |a(u,v)|.\end{align}
$(*)$ Here we apply this inequality for $p=2$.
Alternatively, note that
$$\|u_x\|^2_2\leq \|u\|_2^2+\|u_x\|^2_2+\|u_y\|_2^2=\|u\|_{H^1}^2$$
and thus $\|u_x\|_2\leq \|u\|_{H^1}$. A similar estimate is valid for the other terms in $(1)$. Therefore
$$ a(u,v)| \leq \|u\|_{H^1} \|v\|_{H^1} + \|u\|_{H^1} \|v\|_{H^1} + a \|u\|_{H^1} \|v\|_{H^1}
=(1+1+a)\|u\|_{H^1} \|v\|_{H^1}.$$
