= Solution
{c}
The <Sobolev space> $H^1(b,c)$ consists of $L^2$ functions with a weak first derivative in $L^2$, with norm
$$
\|v\|_{H^1}^2=\int_b^c(|v|^2+|v'|^2)\,dx.
$$
In one dimension these functions have absolutely continuous representatives and well-defined endpoint <Sobolev traces>. Put
$$
V=\{v\in H^1(b,c):v(b)=0\}.
$$
This is a closed subspace, hence a Hilbert space. Both trial and test functions belong to $V$; one does not impose a derivative boundary condition in this space.
Integrate the differential equation against $v\in V$:
$$
\int_b^cfv=\int_b^ce^xu'v'\,dx-e^cu'(c)v(c).
$$
The printed condition uses the ordinary derivative, $u'(c)=-u(c)$, so the weak form is
$$
\boxed{a(u,v)=\int_b^ce^xu'v'\,dx+e^cu(c)v(c),\qquad
\ell(v)=\int_b^cfv\,dx.}
$$
Seek $u\in V$ with $a(u,v)=\ell(v)$ for all $v\in V$. The coefficient $e^c$ in the endpoint term must be retained: replacing the ordinary derivative with the conormal flux would change the boundary problem.
Let $L=c-b$. The <Poincaré inequality> on $V$ can be stated as $\|v\|_{L^2}\leq L\|v'\|_{L^2}$. Also $|v(c)|\leq\sqrt L\|v'\|_{L^2}$. Consequently
$$
a(v,v)\geq e^b\|v'\|_{L^2}^2
\geq\alpha\|v\|_{H^1}^2,\qquad
\alpha=\frac{e^b}{1+L^2}>0,
$$
proving coercivity. By <Cauchy-Schwarz inequality> and the trace bound,
$$
|a(u,v)|\leq e^c(1+L)\|u\|_{H^1}\|v\|_{H^1}.
$$
Thus continuity holds with $M=e^c(1+L)$. The load is continuous if $f\in L^2$, and the <Lax-Milgram theorem> gives a unique weak solution.
Choose a finite-dimensional conforming subspace $V_h\subset V$, such as continuous piecewise linear functions vanishing at $b$, and solve $a(u_h,v_h)=\ell(v_h)$ for every $v_h\in V_h$. Subtracting the weak equations gives <Galerkin orthogonality> $a(u-u_h,v_h)=0$.
<Céa's lemma> states, for a continuous coercive bilinear form with constants $M,\alpha$ on a Hilbert space,
$$
\boxed{\|u-u_h\|_{H^1}\leq\frac M\alpha
\inf_{v_h\in V_h}\|u-v_h\|_{H^1}.}
$$
To prove it, put $e=u-u_h$ and use orthogonality to write $a(e,e)=a(e,u-v_h)$ for any $v_h\in V_h$. Coercivity and continuity give $\alpha\|e\|^2\leq M\|e\|\|u-v_h\|$. If $e=0$ the claim is immediate; otherwise cancel its norm and take the infimum. Any approximation theorem making the infimum tend to zero under mesh refinement therefore gives strong convergence $u_h\to u$ in $H^1$.
Finally, $a$ is symmetric and positive definite on $V$. It is an inner product, so $\|v\|_a=\sqrt{a(v,v)}$ is an <energy norm> on all of $V$: homogeneity and definiteness are immediate, and <Cauchy-Schwarz inequality> for this inner product gives the triangle inequality. No smaller space is required. In this norm the continuity and coercivity constants are one. More precisely, <Galerkin orthogonality> gives
$$
\|u-v_h\|_a^2=\|u-u_h\|_a^2+\|u_h-v_h\|_a^2,
$$
so
$$
\boxed{\|u-u_h\|_a=\min_{v_h\in V_h}\|u-v_h\|_a.}
$$
The finite element solution is the exact best energy-norm approximation.
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