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I am trying to understand why the Lax Milgram theorem is a consequence of the following theorem:

Let $H_1, H_2$ be Hilbert spaces and $k : H_1 \times H_2 \rightarrow \mathbb{R}$ be a continuous bilinear form. For $f \in H_2'$ consider the variational problem: \begin{equation} \text{find } u \in H_1 \text{ such that } k(u,v) = f(v) \text{ for all }v ∈ H_2. \end{equation} The following two statements are equivalent:

  1. For arbitrary $f \in H_2'$ the problem above has a unique solution $u \in H_1$ and $∥u∥_{H_1} ≤ c∥f∥_{H_2'}$ holds with a constant $c$ independent of $f$.

  2. The conditions \begin{equation} \exists \varepsilon >0 : \sup_{v \in H_2}{\frac{k(u,v)}{∥v∥_{H_2}}}\geq \varepsilon ∥u∥_{H_1} \text{ for all } u \in H_1, \end{equation} \begin{equation} \forall v \in H_2, \quad v\neq 0, \quad \exists u \in H_1: \quad k(u,v) \neq 0 \end{equation} hold.

Moreover, for the constants $c$ and $\varepsilon$ one can take $c = \frac{1}{\varepsilon}$ .

Any help is much appreciated.

mkfoi
  • 35

1 Answers1

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The Lax-Milgram theorem is stated once again for convenience.

Let $T$ be a bounded coercive bilinear form on a Hilbert space $H$, with coercive constant $C_c>0$ i.e. for all $u \in H, u \neq 0$ we have $T(u,u) \geq C_c\|u\|_H^2$. Then for all $f \in H'$ there exists $u_f \in H$ such that $\|u_f\|_H \leq \frac{\|f\|_{H'}}{C_c}$ and $T(u_f,v) = f(v)$ for all $v \in H$.

Proof : Let $H_1 = H_2 = H$. We will show that the condition $2$ holds in the theorem given, which will imply condition $1$.

For condition $2$ to hold, we need to show that there is an $\epsilon > 0$ such that for all $u \in H$ we have $\sup_{v \neq 0 \in H} \frac{T(u,v)}{\|v\|_{H}} \geq \epsilon \|u\|_H$. However, note that for $u=0$ the condition is trivially satisfied for any $\epsilon$, and for $u \neq 0$ we have : $$ \sup_{v \neq 0 \in H} \frac{T(u,v)}{\|v\|_{H}} \geq \frac{T(u,u)}{\|u\|_{H}} \geq C_c \|u\|_H $$ hence the first part holds with $\epsilon = C_c$. For the second part, if $v \neq 0 \in H$ then $T(v,v) \geq C_c \|v\|_H^2 > 0$, so $u = v$ is such that $T(u,v) \neq 0$. Thus, condition $2$ holds.

By the theorem given, condition $1$ holds. Condition $1$ says : for all $f \in H'$ there is a unique $u_f \in H$ such that $\|u_f\|_{H} \leq \frac{\|f\|_{H'}}{C_c}$ and $T(u,v) = f(v)$ for all $v \in H$. But this is precisely the Lax-Milgram theorem!


Note that the second theorem is much stronger for a couple of reasons : one is the fact that it deals with forms which are not just bilinear but also takes arguments from different spaces. This allows us to experiment with PDE where arguments from different spaces (for example in PDE, different Sobolev spaces for two functions which have been differentiated a different number of times) are involved. The second improvement is that this helped tackle boundary problems in PDE where the domain "changes shape" with time, the best example being that of a melting iceberg where of course with time the shape of the iceberg changes according to how heat dissipates in the iceberg, so basically the heat equation is solved with shape changes in the domain.

The second theorem is referred to as the Lions-Lax-Milgram theorem, or just Lions' theorem as well, and is a very vital improvement. Another is the Babushka-Brezzi theorem, these are all fundamental for existence and uniqueness theorems for PDE. More on this can be found in Kesavan's book 'Functional analysis and applications'.