Brezis book, Functional analysis, Sobolev spaces and PDE, problem 8.30 Let $k \in \mathbb{R}$, $k \neq 1$, consider the space $$V= \{ v \in H^1(0,1): v(0) = k v(1)\}$$ and the bilinear form $$B(u,v) = \int_0^1 \left(u'v' + uv\right) ~dx - \left(\int_0^1 u\right) \left(\int_0^1 v\right).$$
Now, for above bilinear form, I have to prove it is continuous and coercive. I was able to the show  the continuity and stuck in proving the coercivity? Any leads or ideas are appreciated.
Hint is given: Show for $v \in V$, $\|v\|_{L^\infty(0,1)} \leq C \|v'\|_{L^2(0,1)}$
 A: Proof of the Hint: By Theorem 8.2, we can assume that $v$ is continuous. Then,
$$(k-1)v(0)=v(1)-v(0)=\int_{0}^{1}v'.$$
This implies that $|v(0)|\leq |k-1|^{-1}\|v'\|_{L^2}$. Now, for any $x>0$ in the interval
$$v(x)-v(0)=\int_{0}^{x}v'.$$
Then,
$$|v(x)| \leq |v(0)|+\int_{0}^{x}|v'| \leq |v(0)|+\int_{0}^{1}|v'| \leq C \|v'\|_{L^2},$$
where $C=1+|k-1|^{-1}$. Observe that $C>1$.
Coercivity: Take $\varepsilon \in (0,1)$ so that $\varepsilon C^{2}<1/2$ (e.g. $\varepsilon=1/(4C^{2})$). Then
\begin{align*}
B(v,v) &=\int_{0}^{1} ((v')^{2}+v^{2})dx-\left( \int_{0}^{1}v \right)^{2}  \\ & \geq  \int_{0}^{1} ((v')^{2}+v^{2})dx- \int_{0}^{1}v^{2} \\
&=\int_{0}^{1} ((v')^{2}+v^{2})dx- \varepsilon \int_{0}^{1}v^{2} -(1- \varepsilon)\int_{0}^{1}v^{2} \\
&=\left(\int_{0}^{1}[(v')^{2}-\varepsilon v^{2}]dx \right)+ \left(\int_{0}^{1}v^{2}dx-(1-\varepsilon) \int_{0}^{1}v^{2}dx  \right) \\
&=\left(\int_{0}^{1}[(v')^{2}-\varepsilon v^{2}]dx \right)+\varepsilon \int_{0}^{1}v^{2}.
\end{align*}
Observe that
$$\int_{0}^{1}\varepsilon v^{2}dx \leq \varepsilon \|v\|_{L^{\infty}}^{2} \leq \varepsilon C^{2} \|v'\|_{L^{2}}^{2}<\frac{1}{2}\|v'\|_{L^{2}}^{2}.$$
Putting all the inequalities together, we find
$$B(v,v) \geq \frac{1}{2} \int_{0}^{1}(v')^{2}+\varepsilon \int_{0}^{1}v^{2} \geq \min\left\lbrace \frac{1}{2},\varepsilon \right\rbrace \|v\|_{H^{1}}.$$
