= Solution
The real <Lax-Milgram theorem> states that if $V$ is a <Hilbert space>, $B:V\times V\to\mathbb R$ is a continuous bilinear form with $B(v,v)\ge\alpha\|v\|_V^2$ for some $\alpha>0$, and $\ell\in V^*$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell(v)$ for all $v\in V$. Moreover $\|u\|_V\le\|\ell\|/\alpha$.
Take the <Sobolev space> $V=H_0^1(\Omega)$ with norm $\|v\|_V=\|\nabla v\|_2$. Since $\Omega$ is bounded, the <Poincaré inequality> gives $\|v\|_2\le C_P\|\nabla v\|_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient define
$$
B(u,v)=\int_\Omega(\nabla u\cdot\nabla v+a(x)uv)dx,\qquad\ell(v)=\int_\Omega fv\,dx.
$$
Then $|B(u,v)|\le(1+\bar a C_P^2)\|u\|_V\|v\|_V$, $B(v,v)\ge\|v\|_V^2$ since $a\ge0$, and $|\ell(v)|\le C_P\|f\|_2\|v\|_V$. The <Lax-Milgram theorem> therefore gives a unique $u\in H_0^1(\Omega)$ satisfying this identity, which is exactly the required <weak solution>. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.
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