Show that $L(v) = \int_{\Gamma} gv ds$ is a continuous operator This is problem 2.4 from "Numerical Solution of Partial Differential Equations by the Finite Element Method" by Claes Johnson.
Let $\Omega$ be a square with boundary $\Gamma$. Show that there is a constant C such that
$$\left(\int_\Gamma v^2 ds\right)^{1/2}\le C||v||_{H^1(\Omega)}, \forall v\in H^1(\Omega)$$
Then use that result to show that
$$L(v)=\int_\Gamma gv ds$$
is continuous if $g\in L^2(\Gamma)$.
Idea: I think I need to use Green's theorem, and the fact that were are specifically working with a square. However, I don't know how to turn that into an actual mathematical argument.
 A: The first inequality is indeed Trace theorem (see Evans PDE page 272). The idea is as follows:
    Suppose $\Omega=[0,1]\times[0,1]$, and $\Gamma = \Gamma_1\cup \Gamma_2\cup \Gamma_3\cup \Gamma_4$, where
    \begin{align*}
  \Gamma_1&=[0,1]\times \left\{ 0 \right\};\\
  \Gamma_2&=[0,1]\times \left\{ 1 \right\};\\
  \Gamma_3& = \left\{ 0 \right\}\times[0,1];\\
  \Gamma_4& = \left\{ 1 \right\}\times[0,1].
 \end{align*}
    Then select a cut off function $\zeta\in C^{\infty}(\Omega)$ s.t. $\zeta=1$ on $\Gamma_1$ and $\zeta=0$ on $\Gamma_2$. 
    \begin{align*}
  \int_{\Gamma_1}v^2ds &= -\int_{\Omega}(\zeta v^2)_{x_2}dx_1dx_2\\
  & =-\int_{\Omega}\zeta_{x_2}v^2+\zeta vv_{x_2}dx_1dx_2\\
  & \leq C\int_{\Omega}|v|^2+|Dv|^2dx_1dx_2
 \end{align*}
    where we used Young's inequality. In a similar way, one can show that this inequality holds for $\Gamma_2$, $\Gamma_3$ and $\Gamma_4$. In all, 
    \begin{equation}
  \left( \int_{\Gamma}v^2ds \right)^{\frac{1}{2}}\leq C\|v\|_{H^{1}(\Omega)}.
 \end{equation}
    For the second part, since $L$ is a linear funciotnal over $H^{1}(\Omega)$, therefore it is continuous if and only if it is bounded, that is
    \begin{equation}
  L(v)=\int_{\Gamma}^{}gvds\leq \|g\|_{L^2(\Gamma)}\|v\|_{L^2(\Gamma)}\leq C\|v\|_{H^1(\Omega)}.
 \end{equation}
