Let $I$ be the open interval $(0, 1)$. Assume that $p \in C^1(\bar I)$ and $q \in$ $C(\bar I)$ such that $p(x) \geq \alpha>0$ and $q(x) \geq 0$ for all $x \in \bar I$. Let $f \in L^2 (I)$.
In the solution for exercise 8.21.2 in his book Functional Analysis, Brezis considers the equation $$ (*) \quad \begin{cases} -(pU')' + qU &= f \text{ on } I, \\ U(1) &= 0. \end{cases} $$
I tried to use Lax–Milgram theorem but could not obtain the continuity and coercivity of the bilinear form $a$. Could you elaborate on how to prove the existence of above ODE?
My attempt If $U$ is a classical solution of $(*)$, then $$ -\int_I (pU')' V + \int_I qUV= \int_I fV, \quad \forall V \in H^1(I). $$
Let $\mathcal K := \{V \in H^1 (I): V(1)=0\}$. Then $V$ is a closed subspace of $H^1 (I)$. By integration by parts, for each $V \in \mathcal K$ we have $$ \begin{align*} \int_I (pU')' V &= (pU' V) (1) - (pU' V) (0) - \int_I pU'V' \\ &= -(pU' V) (0) - \int_I pU'V'. \end{align*} $$
Then $U$ satisfies $$ (pU' V) (0) + \int_I [qUV + pU'V']= \int_I fV, \quad \forall V \in \mathcal K. $$
We define a (non-symmetric) bilinear form $a$ on $\mathcal K$ by $$ a(U, V) := (pU' V) (0) + \int_I [qUV + pU'V']. $$
We fix an arbitrary $\alpha \in \mathbb R$ and consider $$ (1) \quad \begin{cases} -(pU')' + qU &= f \text{ on } I, \\ U(0) &= \alpha, \\ U(1) &= 0. \end{cases} $$
We fix $g \in C^\infty (\bar I)$ such that $g(0) = \alpha$ and $g(1)=0$. Then $(1)$ is equivalent to $$ (2) \quad \begin{cases} -(p(W+g)')' + q(W+g) &= f \text{ on } I, \\ W(0) &= 0, \\ W(1) &= 0, \end{cases} $$
which is in turn equivalent to $$ (3) \quad \begin{cases} -(pW')' + qW &= h:=f+(pg')'-qg \text{ on } I, \\ W(0) &= 0, \\ W(1) &= 0. \end{cases} $$
Given the hypothesis about $p$ and $q$, we can apply Lax-Milgram theorem (as in your thread) to get the existence of the unique solution $\bar W \in H^2 (I)$ to $(3)$. We get back the unique solution $\bar U \in H^2 (I)$ to $(1)$ by setting $\bar U :=\bar W + g$.