 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 $g_{+}\ge g$, proving that $M$ is nonempty.

If $v,w\in M$ and $t\in [0,1]$, then $(1-t)v+tw\ge g$, so $M$ is a convex set. If $v_n\to v$ in $H_0^1$, the estimate $\Vert v_n-v{\Vert }_{\infty }\le \sqrt{l}\Vert v_n^{\prime }-v^{\prime }{\Vert }_2$ gives uniform convergence. Passing to the limit in $v_n(x)\ge g(x)$ shows $v\in M$. Thus the admissible set is nonempty, closed and convex.