能否利用Lax-Milgram定理证明Stampacchia定理？

First, let's recap the key definitions and theorems we're working with to set the stage:

Background Definitions #

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space with its induced norm $|\cdot|$. A bilinear form $[\cdot, \cdot]:H \times H \to \mathbb{R}$ has two critical properties we care about:

• Continuous: There exists a constant $C>0$ such that
  $$
  |[u, v]| \le C |u| |v| \quad \forall u,v \in H.
  $$
• Coercive: There exists a constant $\alpha>0$ such that
  $$
  [v, v] \ge \alpha |v|^2 \quad \forall v \in H.
  $$

Stampacchia Theorem #

Assume $[\cdot, \cdot]$ is continuous and coercive, and let $K$ be a non-empty closed convex subset of $H$. For every $\varphi \in H^*$, there exists a unique $u \in K$ such that
$$
[u, v-u] \ge \varphi(v-u) \quad \forall v \in K.
$$
If $[\cdot, \cdot]$ is symmetric, $u$ is also the unique minimizer of the functional:
$$
\min_{v\in K} \bigg { \frac{1}{2} [v, v] - \varphi(v) \bigg }.
$$

Lax-Milgram Theorem #

Assume $[\cdot, \cdot]$ is continuous and coercive. For every $\varphi \in H^*$, there exists a unique $u \in H$ such that
$$
[u, v] = \varphi(v) \quad \forall v \in H.
$$
If $[\cdot, \cdot]$ is symmetric, $u$ is the unique minimizer of:
$$
\min_{v\in H} \bigg { \frac{1}{2} [v, v] - \varphi(v) \bigg }.
$$

We already know proving Lax-Milgram via Stampacchia is straightforward—but can we reverse the logic? Yes, absolutely! Here's how to do it, using the projection operator onto closed convex sets as the key link:

Step 1: Define an equivalent Hilbert space structure #

Since $[\cdot, \cdot]$ is continuous and coercive, it defines an equivalent inner product on $H$:
$$
\langle u, v \rangle_A = [u, v] \quad \forall u, v \in H.
$$
Coercivity ensures the norm $|\cdot|_A = \sqrt{[\cdot, \cdot]}$ is equivalent to the original norm $|\cdot|$, so $H$ is still a Hilbert space when equipped with $\langle \cdot, \cdot \rangle_A$.

Step 2: Use Lax-Milgram to find a "representative" element #

By the Lax-Milgram theorem, for our given $\varphi \in H^*$, there exists a unique $z_0 \in H$ such that
$$
[z_0, v] = \varphi(v) \quad \forall v \in H.
$$
This $z_0$ is essentially the Riesz representer of $\varphi$ with respect to the $\langle \cdot, \cdot \rangle_A$ inner product.

Step 3: Project onto the convex set $K$ #

In any Hilbert space, the orthogonal projection onto a non-empty closed convex subset is well-defined. Let $u = P_K^A z_0$, where $P_K^A$ is the projection of $z_0$ onto $K$ in the $\langle \cdot, \cdot \rangle_A$ Hilbert space. By the projection property, we have:
$$
\langle u - z_0, v - u \rangle_A \ge 0 \quad \forall v \in K.
$$
Substituting back the definition of $\langle \cdot, \cdot \rangle_A$, this becomes:
$$
[u - z_0, v - u] \ge 0 \quad \forall v \in K.
$$

Step 4: Rearrange to get the Stampacchia inequality #

Expand the left-hand side:
$$
[u, v - u] - [z_0, v - u] \ge 0.
$$
But from Step 2, $[z_0, v - u] = \varphi(v - u)$, so substituting that in gives:
$$
[u, v - u] \ge \varphi(v - u) \quad \forall v \in K.
$$
Perfect—this is exactly the variational inequality from the Stampacchia theorem!

Step 5: Prove uniqueness #

Suppose there are two elements $u_1, u_2 \in K$ that satisfy the Stampacchia inequality. Then:
$$
[u_1, u_2 - u_1] \ge \varphi(u_2 - u_1)
$$
$$
[u_2, u_1 - u_2] \ge \varphi(u_1 - u_2)
$$
Add these two inequalities together:
$$
[u_1, u_2 - u_1] + [u_2, u_1 - u_2] \ge \varphi(u_2 - u_1) + \varphi(u_1 - u_2).
$$
The right-hand side is zero because $\varphi$ is linear. For the left-hand side, rewrite it as:
$$
[u_1 - u_2, u_2 - u_1] = -[u_1 - u_2, u_1 - u_2] \ge 0.
$$
By coercivity, $[u_1 - u_2, u_1 - u_2] \ge \alpha |u_1 - u_2|^2$, so:
$$
-\alpha |u_1 - u_2|^2 \ge 0 \implies |u_1 - u_2| = 0 \implies u_1 = u_2.
$$
Uniqueness is confirmed!

So there you have it—we've successfully proven the Stampacchia theorem using the Lax-Milgram theorem, with a little help from projection operators in equivalent Hilbert spaces.

备注：内容来源于stack exchange，提问作者Analyst