Consider the Hilbert space $H_0^1(\mathbb R^n)$ defined as the closure of $C^\infty_c(\mathbb R^n)$ in $H^1(\mathbb R^n)$. I am interested in the following convex subspace: $$\mathscr C \,:= \left\{\,u\in H_0^1(\mathbb R^n) \ \Bigg|\ u\geq0\,,\, \int_{\mathbb R^n} u(x)\,dx=1 \,\right\} \,.$$
Part 1
In general Poincaré inequality does not hold on $H_0^1(\mathbb R^n)$. However is it possible to state it on $\mathscr C$? To be precise, is there a constant $c<\infty$ such that $$ ||u||_{L^2(\mathbb R^n)} \,\leq c\,||\nabla u||_{L^2(\mathbb R^n)} \quad \forall\,u\in\mathscr C \quad?$$
Part 2
Is it possible to state a Lax-Milgram theorem for the space $\mathscr C$ ? Something like the following. Let $A(\cdot,\cdot)$ and $L(\cdot)$ be respectively a bilinear and linear form on $H_0^1(\mathbb R^n)$, such that
- A is bounded on $H_0^1(\mathbb R^n)\,$: there exists $C$ such that $A(u,v)\leq C\,||u||_{H_0^1(\mathbb R^n)}\,||v||_{H_0^1(\mathbb R^n)}\,$ for all $u,v\in H_0^1(\mathbb R^n)$
- A is coercive on $\mathscr C\,$: there exists $C$ such that $A(u,u)\geq C\,||u||_{H_0^1(\mathbb R^n)}\,$ for all $u\in\mathscr C$
- L is bounded on $H_0^1(\mathbb R^n)\,$: there exists $C$ such that $L(u)\leq C\,||u||_{H_0^1(\mathbb R^n)}\,$ for all $u\in H_0^1(\mathbb R^n)$
Then given $v\in H_0^1(\mathbb R^n)$ (or possibly in $\mathscr C$), is there a unique solution $u\in\mathscr C$ to the equation $$A(u,v) = L(v) \quad?$$
Part 1 is not true. Take $n=1$, and define $u_k$ to be continuous with $u_k=\frac1{k+1}$ on $[0,k]$, $u=0$ outside of $[-1,k+1]$, affine linear on $[-1,0]$ and $[k,k+1]$. So these functions are piecewise linear bumps with support growing wider for growing $k$.
If I am not mistaken $\int u_k=1$, $\|u_k\|_{L^2}^2 \sim \frac1k$, $\|\nabla u_k\|_{L^2}^2 \sim \frac1{k^2}$, so that $$ \frac{ \|\nabla u_k\|_{L^2} }{\|u_k\|_{L^2} }\to 0. $$
Part 2 can also be answered in the negative. Define $L$ by $Lv=\int lv$ for some negative $l$ with $supp\ l = \mathbb R^n$. Then there cannot be a solution of the problem in $C$ if $A$ equals the $H^1$-inner product.
If one formulates the problem as variational inequality: Find $u\in C$ such that $$ A(u,v-u) \ge L(v-u) \quad \forall v\in C, $$ then this is always uniquely solvable. The corresponding theorem reduces to Lax-Milgram if $C=H^1$.