Lax-Milgram Theorem

Lax-Milgram Theorem was named after Peter Lax (1926 - 2025) and Arthur Milgram (1912 - 1961), who proved it in 1954. It is stated as follows. Initially, we are given a real Hilbert space H. Denote by R the set of real numbers. We consider a bilinear mapping B: H x H -> R satisfying the following two properties:

   (i) |B[u, v]| <= a.||u||.||v|| for all u, v belonging to H, 

   (ii) b||u||^2 <= B[u, u] for all u belonging to H, 

where a, b are two positive numbers that doesn't depend on u, v. In other words, each such mapping B is associated with a pair (a, b) (trivially b <= a). Finally, let f: H -> R be a bounded linear functional on H. Then there exists a unique element u belonging to H such that B[u, v] = <f, v> for all v belonging to H. 

Different books might have different ways to describe the proof of this theorem. The proof that is presented here can be found in the book "Partial Differential Equations" written by Lawrence C. Evans. It is divided into six parts. The least technical (and most impressive, from my own perspective) parts are, probably, the third and the fourth part. We will focus on these two parts, while the other parts are just simply presented without further explanations. 

   1. Fix an element u and consider the mapping v -> B[u, v]. By the Riesz Representation Theorem, there exists w (depending only on u) such that B[u, v] = (w, v). We define a new mapping A(u) = w. 

   2. By technical argument, we can prove that A is linear. Moreover, ||A(u)||^2 = (A(u), A(u)) = B[u, A(u)] <= a*||u||*||A(u)|| and ||A(u)|| <= a*||u||, so A is bounded. 

   3. Amazingly, in the next step, we prove that A is injective and the set of all possible values of A(u) (called the range of A, denoted by R(A)) is closed in H. We have b*||u||^2 <= B[u, u] = (A(u), u) <= ||A(u)||*||u||, which implies b*||u|| <= ||A(u)||. It was written in the book that "this inequality easily implies (4)", where (4) denoted the fact that A is injective and R(A) is closed in H. I will give a clearer explanation. First, consider a sequence A(u_1), A(u_2), ..., A(u_k), ... of elements of R(A) that converges to an element f. Since H is a Hilbert space, every Cauchy sequence is convergent. A(u_1), A(u_2), ..., A(u_k) is definitely a Cauchy sequence and for every i, j, ||u_i - u_j|| <= A(u_i - u_j)/b, thus u_1, u_2, ..., u_k, ... is also a Cauchy sequence. This sequence converges to an element u. By 2., A is bounded and continuous, thus A(u_k) converges to A(u) and f = A(u). Therefore, f belongs to R(A) and R(A) is closed in H. The injectivity of A is more obvious. If A(u) = 0, then ||A(u)|| = 0 and b*||u|| = 0, which implies ||u|| = 0 and u = 0.   

   4. We prove that R(A) = H. First, recall from Functional Analysis that if M is closed and M is not H itself, then there exists a nonzero element that is perpendicular to all elements of M. As a result, if R(A) is not H itself, then there exists a nonzero element w that is perpendicular to all elements of R(A). Since A(w) is obviously an element of R(A), w is also perpendicular to A(w). This means that (A(w), w) = 0, which is absurd since (A(w), w) = B[w, w] >= b*||w||^2 > 0. 

   5. Again, by Riesz Representation Theorem, we can choose w such that <f, v> = (w, v) for all v belonging to H. Since R(A) = H, w can be represented as A(u), where u is an element of H. As a result, B[u, v] = (A(u), v) = (w, v) = <f, v>, which is the desired "equation". 

   6. The final step is to prove the uniqueness of u. If there are two such u, namely u(1) and u(2), then B[u(1) - u(2), v] = 0, which is again a contradiction by the condition (ii).