Let $I$ be the open interval $(0, 1)$. Let $k \in \mathbb R$ such that $k \neq 1$. We consider the space $$ V := \{v \in H^1 (I) : v(0) = kv(1)\}, $$ and the symmetric bilinear form $a$ defined on $V$ by $$ a(u, v) = \int_I [ u'v' + uv ] - \left ( \int_I u \right) \left ( \int_I v \right). $$
I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.30
- Check that $V$ is a closed subspace of $H^1 (I)$.
In what follows, $V$ is equipped with the Hilbert structure induced by the $H^1$ inner product.
- Prove that $a$ is a continuous and coercive bilinear form on $V$.
There are possibly subtle mistakes that I could not recognize in my below attempt of (2.). Could you please have a check on it?
The continuity of $a$ is easy. Let's prove the coercivity of $a$. We need to find $\varepsilon >0$ such that $$ a(u, u) = \int_I [ |u'|^2 + |u|^2 ] - \left ( \int_I u \right)^2 \ge \varepsilon \|u \|_{H^1}^2 = \varepsilon \int_I [ |u'|^2 + |u|^2 ], $$ which is equivalent to $$ (1-\varepsilon) \int_I [ |u'|^2 + |u|^2 ] \ge \left ( \int_I u \right)^2. $$
We have $\left ( \int_I u \right)^2 \le \int_I |u|^2$, so it suffices to find $\varepsilon >0$ such that $$ (1-\varepsilon) \int_I [ |u'|^2 + |u|^2 ] \ge \int_I |u|^2, $$ which is equivalent to $$ \int_I |u'|^2 \ge \frac{\varepsilon}{1-\varepsilon} \int_I |u|^2. $$
For $x \in I$, we have $u(x) - u(0) = \int_0^x u'$ and $u(1) - u(x) = \int_x^1 u'$. We have $ku(1)-u(0)=0$, so $(1-k) u(x) = \int_0^x u' + k \int_x^1 u'$. Then $$ |u(x)| \le \frac{k+1}{|k-1|} \int_I |u'|, $$ which implies $$ \int_I |u|^2 \le \left ( \frac{k+1}{k-1} \right )^2 \int_I |u'|^2. $$
We just need to pick $\varepsilon >0$ such that $$ \left ( \frac{k+1}{k-1} \right )^2 \le \frac{1- \varepsilon}{\varepsilon}. $$
