Lax-Milgram as a corollary of Stampacchia theorem [Brezis book]

Question

I'm reading Brezis functional analysis,sobolev spaces and pdes book, where at page 140 of the 2010 Springer edition there is the following corollary 5.8, that is a corollary of Stampacchia theorem (theorem 5.6.):

Assume $a(u,u)$ is a coercive bilinear form on $H$ (where $H$ is a Hilbert space over $\mathbb{R}$); then for every $\phi \in H^*$ there exists an element $u$ such that $a(u,v)=\langle \phi,v \rangle,$ for every $v \in H.$ Moreover, if $a$ is symmetric, $u$ is characterized by the property $\mathcal{P}:$ $$u\in H \ \ \text{and} \ \ \frac{1}{2}a(u,u)-\langle \phi,u \rangle = \text{min}_{v \in H}\big\{\frac{1}{2} a(v,v)-\langle \phi,v \rangle \big\}$$

Unfortunately, Brezis gives a very sketchy proof of the corollary, saying that one should just apply the reasoning of a previous corollary (5.4), which say that

if $M$ is a closed linear subspace of $H.$ For $x\in H,$ $y=P_Kx$ is characterized by the property that for all $m \in M$ $$ y\in M \ \text{and} \ \langle x-y, m \rangle =0$$

question

Can you provide me a more detailed proof of this fact? Either a proof given as an answer or a reference to a detailed proof in some other book is good.

Plese read

The only request is that I would like to follow the approach of Brezis of deriving it from Stampacchia theorem.

Answer 1

Firstly, consider the linear operator $A:H\to H^{*}$ defined by $Au(v)=a(u,v)$. Then by the coercivity exist $C>0$ such that $||A^{*}v||\geq C||v||$. Theorem 2.22 from Brezis in page 47 implies $A$ is a surjective operator. For all $\phi \in V^{*}$, exist $u \in H$ with $Au=\phi$. Observe that coercivity says $A$ in injective so $A$ is bijective.

For second claim define the "energy" $$E(u)=\frac{1}{2}a(u,u)-\langle \phi,u\rangle.$$ Bound by below you obtain $E(u)\geq \frac{C}{2}||u||²-||\phi||\cdot ||u|| \geq \frac{1}{2C}(C||u||-||\phi||)²-\frac{1}{2c}||\phi||²$, then infimun exist. Take a sequence $E(u_n) \to \inf_{v \in H} E(v)$ and prove $u_n$ is a Cauchy sequence in $H$ (use coercivity). $H$ is hilbert space so the minimizer sequence is a minimum.

Answer 2

To see he second claim, define $$ E(u) = \frac12 a(u,u) - (\phi,u). $$ Then for $u,v$ it holds $$ E(u+v) - E(u) = a(u,v)-(\phi,v) + a(v,v). $$ This shows immediately, that if $a(u,v)-(\phi,v)=0$ for all $v$, then $u$ minimizes the energy. Assume now that $u$ minimizes the enery, then we have for all $t>0$ and $v$ $$ 0\le E(u+tv) - E(u) = t(a(u,v)-(\phi,v)) + t^2a(v,v). $$ Dividing by $t$, and passing to the limit $t\searrow0$, we obtain $$ a(u,v)-(\phi,v) \ge0 \quad \forall v. $$ This equally is true for $-v$, so $a(u,v)-(\phi,v) =0$ for all $v$.