Unique solution of a symmetric coercive variational problem on a Hilbert space

For every $v\in H$ one has $\frac{1}{2}a(v,v)\ge \frac{\alpha }{2}\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by the coercivity hypothesis and $-\mathrm{\ell }(v)\ge -\mathrm{\mid }\mathrm{\ell }(v)\mathrm{\mid }\ge -\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ by [F4], hence $J(v)\ge \frac{\alpha }{2}\mathrm{\parallel }v{\mathrm{\parallel }}^2-\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$.

Expanding with bilinearity and symmetry of $a$ (the same expansion as the parallelogram identity [F2], applied to the quadratic form $v\mapsto a(v,v)$) gives $a(\frac{v+w}{2},\frac{v+w}{2})=\frac{1}{2}a(v,v)+\frac{1}{2}a(w,w)-\frac{1}{4}a(v-w,v-w)$, hence $J\left(\frac{v+w}{2}\right)=\frac{J(v)+J(w)}{2}-\frac{1}{8}a(v-w,v-w)$ for all $v,w\in H$.

Completing the square gives $\frac{\alpha }{2}\mathrm{\parallel }v{\mathrm{\parallel }}^2-\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }=\frac{\alpha }{2}(\mathrm{\parallel }v\mathrm{\parallel }-\frac{\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }}{\alpha }{)}^2-\frac{\mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}^2}{2\alpha }\ge -\frac{\mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}^2}{2\alpha }$, so the image $S=\{J(v):v\in H\}$ is nonempty and bounded below in $\mathbb{R}$, and its infimum $m=\inf S\in \mathbb{R}$ exists by [F5].

For each $n\in \mathbb{N}$ the set $X_n:=\{v\in H:J(v)<m+\frac{1}{n+1}\}$ is nonempty: otherwise $m+\frac{1}{n+1}$ would be a lower bound of $S$ larger than $m$, contradicting [F5] (equivalently, [F5] with $\epsilon =\frac{1}{n+1}$ produces $v$ with $J(v)<m+\frac{1}{n+1}$). Applying ${\mathrm{A}\mathrm{C}}_{\omega }$ [F6] to the family $(X_n{)}_{n\in \mathbb{N}}$ gives a sequence $(v_n)$ in $H$ with $J(v_n)<m+\frac{1}{n+1}$ for all $n$; since $m$ is a lower bound of $S$, also $J(v_n)\ge m$, so $J(v_n)\to m$.

Since $m$ is a lower bound of $S$ one has $J\left(\frac{v_n+v_k}{2}\right)\ge m$, so step 1.2 and the coercivity bound $a(v_n-v_k,v_n-v_k)\ge \alpha \mathrm{\parallel }{v}_n-v_k{\mathrm{\parallel }}^2$ give $\frac{\alpha }{8}\mathrm{\parallel }{v}_n-v_k{\mathrm{\parallel }}^2\le \frac{J(v_n)+J(v_k)}{2}-m\le \frac{1}{2}(\frac{1}{n+1}+\frac{1}{k+1})$; the right-hand side tends to $0$ as $n,k\to \mathrm{\infty }$, so $(v_n)$ is a Cauchy sequence in $H$.

By completeness [F1] there is $u\in H$ with $\mathrm{\parallel }{v}_n-u\mathrm{\parallel }\to 0$.

The functional $J$ is continuous in the norm: for $v,w\in H$ one has $\mathrm{\mid }J(v)-J(w)\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }a(v,v)-a(w,w)\mathrm{\mid }+\mathrm{\mid }\mathrm{\ell }(v-w)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }a(v+w,v-w)\mathrm{\mid }+\mathrm{\mid }\mathrm{\ell }(v-w)\mathrm{\mid }\le (\frac{C}{2}\mathrm{\parallel }v+w\mathrm{\parallel }+\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel })\mathrm{\parallel }v-w\mathrm{\parallel }$ by [F3] and [F4], which tends to $0$ as $v\to w$; hence $J(u)={\lim }_{n}J(v_n)=m$.

Fix $v\in H$ and consider $g(s)=J(u+sv)$; expanding with bilinearity and symmetry gives $g(s)=J(u)+s(a(u,v)-\mathrm{\ell }(v))+\frac{s^2}{2}a(v,v)$ for all $s\in \mathbb{R}$. Since $J(u)=m$ is the minimum of $J$, $g(s)\ge g(0)$ for all $s$: for $s>0$ division by $s$ gives $a(u,v)-\mathrm{\ell }(v)+\frac{s}{2}a(v,v)\ge 0$, and letting $s\to 0^{+}$ gives $a(u,v)-\mathrm{\ell }(v)\ge 0$, while for $s<0$ the same division gives $a(u,v)-\mathrm{\ell }(v)+\frac{s}{2}a(v,v)\le 0$, and letting $s\to 0^{-}$ gives $a(u,v)-\mathrm{\ell }(v)\le 0$. Therefore $a(u,v)=\mathrm{\ell }(v)$, and since $v\in H$ was arbitrary, $u$ solves the variational equation.

Uniqueness: if $u_1,u_2\in H$ both satisfy the equation, then testing the difference at $v=u_1-u_2$ gives $a(u_1-u_2,u_1-u_2)=0$, and coercivity gives $\alpha \mathrm{\parallel }{u}_1-u_2{\mathrm{\parallel }}^2\le a(u_1-u_2,u_1-u_2)=0$, so $\mathrm{\parallel }{u}_1-u_2\mathrm{\parallel }=0$ and $u_1=u_2$ by [F1].

Estimate: testing the equation at $v=u$ gives $a(u,u)=\mathrm{\ell }(u)$, and coercivity together with [F4] gives $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2\le a(u,u)=\mathrm{\ell }(u)\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$; if $\mathrm{\parallel }u\mathrm{\parallel }=0$ then $0\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{/}\alpha$ holds, and otherwise division by $\alpha \mathrm{\parallel }u\mathrm{\parallel }>0$ gives $\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\mathrm{/}\alpha$. This proves existence, uniqueness and the estimate, and identifies the sole use of ${\mathrm{A}\mathrm{C}}_{\omega }$ as the selection in step 3.1.