= Solution
One-dimensional <Sobolev embedding> gives a continuous representative for each $v\in H_0^1(0,l)$, with zero endpoint values, so the pointwise obstacle condition is meaningful. The <positive part> $g_+=\max\{g,0\}$ is <Lipschitz> and vanishes near both endpoints because $g(0),g(l)<0$. It therefore belongs to $H_0^1$ and satisfies $g_+\geq g$, proving that $M$ is nonempty.
If $v,w\in M$ and $t\in[0,1]$, then $(1-t)v+tw\geq g$, so $M$ is a <convex set>. If $v_n\to v$ in $H_0^1$, the estimate $\|v_n-v\|_\infty\leq\sqrt l\,\|v_n'-v'\|_2$ gives <uniform convergence>. Passing to the limit in $v_n(x)\geq g(x)$ shows $v\in M$. Thus \b[the admissible set is nonempty, closed and convex].
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