Let $I$ be the open interval $(-1, 1)$. Consider the bilinear form on $H^1_0 (I)$ $$ a(u, v) = \int_I [ u'v' + uv + \lambda u(0) v], $$ where $\lambda \in \mathbb R$ is fixed. I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.27
Check that $a(u, v)$ is a continuous bilinear form on $H_0^1(I)$.
Prove that if $|\lambda|<\sqrt{2}$, the bilinear form $a$ is coercive.
Deduce that if $|\lambda|<\sqrt{2}$, then for every $f \in L^2(I)$ there exists a unique solution $u \in H^2(I) \cap H_0^1(I)$ of the problem $$ (1) \quad \begin{cases} -u'' + u-\lambda u(0)=f \quad \text {on} \quad I, \\ u(-1)=u(1)=0. \end{cases} $$
There are possibly subtle mistakes that I could not recognize in my below attempt of (2.) and (3.). Could you please have a check on it?
We have $|u(0)| = |\int_{-1}^0 u'|$, so $|u(0)|^2 \le \int_I |u'|^2$. We have $$ \begin{align*} \lambda |u(0)| \int_I u &\le \sqrt 2 \lambda |u(0)| \sqrt{\int_I u^2} \quad \text{by Hölder's inequality} \\ &\le \frac{\lambda}{\sqrt 2} \left ( |u(0)|^2 + \int_I u^2 \right ) \quad \text{by AM-GM inequality} \\ &\le \frac{\lambda}{\sqrt 2} \left ( \int_I |u'|^2 + \int_I u^2 \right ), \end{align*} $$ which implies $$ a(u, u) \ge \left ( 1- \frac{\lambda}{\sqrt 2}\right ) \|u\|_{H^1}^2. $$
If $u$ is a classical solution to $(1)$, then $$ (2) \quad \int_I [-u''w + uw -\lambda u(0) w] = \int_I fw, \quad \forall w \in H^1 (I), $$ which (by integration by parts) implies $$ (3) \quad a(u, w) = \int_I fw, \quad \forall w \in H^1_0 (I). $$ By Lax-Milgram, $(3)$ has a unique solution $u \in H^1_0 (I)$. We have also from $(3)$ that $$ \int_I u' w' = - \int_I (u + \lambda u(0) - f) w, \quad \forall w \in H^1_0 (I), $$ which implies $u' \in H^1 (I)$ with $u'' = u + \lambda u(0) - f$. Then $u \in H^2 (I)$.