= Solution
Use a <Lagrange multiplier> $\mu$ for the norm constraint. For real variations $u+\delta\eta$ with $\eta(0)=\eta(\pi)=0$, <integration by parts> gives
$$
\begin{aligned}
\delta(F-\mu G)&=2\int_0^\pi[u'\eta'+(\epsilon x-\mu)u\eta]dx\\
&=2\int_0^\pi[-u''+(\epsilon x-\mu)u]\eta\,dx.
\end{aligned}
$$
Thus stationarity is equivalent to the <Sturm-Liouville problem>
$$
-u''+\epsilon xu=\mu u,\qquad u(0)=u(\pi)=0.
$$
Multiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the <eigenvalues>, and the normalized stationary functions are their <eigenfunctions>. Conversely the eigenvalue equation makes the constrained first variation vanish. The <self-adjoint operator> has weight one.
For $\epsilon=0$, zero and negative eigenvalues yield no nonzero solution with both <Dirichlet boundary conditions>. Positive eigenvalues have solutions $u=A\sin(\sqrt\lambda x)$ after the left condition, and the right condition gives $\sqrt\lambda=n$, $n\geq1$. Hence the lowest eigenpair is
$$
\boxed{\lambda_0=1,\qquad Y_1(x)=\sqrt{\frac2\pi}\sin x,}
$$
where the amplitude follows from $\int_0^\pi\sin^2x\,dx=\pi/2$.
The <Rayleigh-Ritz variational principle> bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi/2$ gives
$$
\int_0^\pi x\sin^2x\,dx=\frac\pi2\int_0^\pi\sin^2x\,dx=\frac{\pi^2}{4}.
$$
Thus the <sine trial bound for a linear potential on an interval> is
$$
\boxed{\lambda_\epsilon\leq1+\frac{\epsilon\pi}{2}.}
$$
For $\epsilon>0$ the bound is strict, since $-Y_1''+\epsilon xY_1=(1+\epsilon x)Y_1$ is not a constant multiple of $Y_1$. Also $\lambda_\epsilon\geq1$ by the <Poincaré inequality> and the nonnegative potential. Equality in the upper bound holds at $\epsilon=0$.
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