= Solution
A real <bilinear form> is a <coercive bilinear form> on $V$ when $a(v,v)\geq\alpha\|v\|_V^2$ for all $v\in V$ and some fixed $\alpha>0$. It is a <bounded bilinear form>, equivalently continuous, when $|a(v,w)|\leq M\|v\|_V\|w\|_V$ for one finite $M$. A continuous <linear functional> satisfies $|\ell(v)|\leq C_\ell\|v\|_V$.
The <Galerkin method> seeks $u_h\in S$ with $a(u_h,v_h)=\ell(v_h)$ for all $v_h\in S$. Subtraction from the continuous equations gives <Galerkin orthogonality>: $a(u-u_h,v_h)=0$ for every $v_h\in S$. Under the stated continuity and coercivity assumptions, <Céa's lemma> says
$$
\boxed{\|u-u_h\|_V\leq\frac M\alpha\inf_{v_h\in S}\|u-v_h\|_V.}
$$
Indeed, with $e=u-u_h$ and any $v_h\in S$, <Galerkin orthogonality> eliminates $v_h-u_h$:
$$
\alpha\|e\|_V^2\leq a(e,e)=a(e,u-v_h)\leq M\|e\|_V\|u-v_h\|_V.
$$
If $e=0$ the claim is immediate; otherwise cancel $\|e\|_V$ and take the <infimum>. No symmetry of $a$ was used. These statements define an approximation when it exists; a finite-dimensional $S$ gives existence and uniqueness by the <Lax-Milgram theorem>, as does a closed <vector subspace>.
For the interval example write $L=c-b$ and $R=\max(|b|,|c|)$. The <one-dimensional Sobolev representative> of $v\in V$ satisfies
$$
v(x)=-\int_x^cv'(t)\,dt,\qquad |v(x)|^2\leq(c-x)\int_x^c|v'(t)|^2\,dt\leq(c-x)\|v'\|_2^2
$$
by the <Cauchy-Schwarz inequality>. The representation follows by taking the <absolutely continuous function> primitive of the <weak derivative>; its difference from $v$ has zero <weak derivative>, hence is constant, and $v(c)=0$ fixes that constant. Integrating the estimate gives the <Endpoint Poincaré inequality>
$$
\|v\|_2^2\leq\frac{L^2}{2}\|v'\|_2^2,\qquad
\|v\|_V^2\leq\left(1+\frac{L^2}{2}\right)\|v'\|_2^2.
$$
Thus the nonnegative potential $x^2$ gives
$$
\boxed{a(v,v)\geq\|v'\|_2^2\geq\frac1{1+L^2/2}\|v\|_V^2.}
$$
For continuity, $|x|\leq R$, the <Cauchy-Schwarz inequality> for the two integrals, and then its two-dimensional version give
$$
|a(u,v)|\leq R^2\|u\|_2\|v\|_2+\|u'\|_2\|v'\|_2\leq\boxed{\max(1,R^2)\|u\|_V\|v\|_V}.
$$
Similarly $\boxed{|\ell(v)|\leq\|f\|_2\|v\|_2\leq\|f\|_2\|v\|_V}$, proving continuity of the <linear functional>.
For the last displayed integral take $\Omega=(-1,1)$ and retain the endpoint condition $v(1)=0$, the domain-consistent interpretation of the changed formula. \b[The changed form is still coercive.] A negative potential on part of the interval does not by itself destroy coercivity. The sharper pointwise estimate above gives
$$
\int_{-1}^0(-x)v(x)^2\,dx\leq\left[\int_{-1}^0(-x)(1-x)\,dx\right]\|v'\|_2^2=\frac56\|v'\|_2^2.
$$
Discarding only the positive part of the potential yields
$$
a(v,v)=\int_{-1}^1(v'^2+xv^2)\,dx\geq\frac16\|v'\|_2^2.
$$
The proved <Endpoint Poincaré inequality> with $L=2$ gives $\|v\|_V^2\leq3\|v'\|_2^2$, and therefore
$$
\boxed{a(v,v)\geq\frac1{18}\|v\|_V^2.}
$$
This is a concrete application of <negative-potential domination by endpoint control>. The constant is sufficient, without claiming to be optimal.
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