# Lax-Milgram theorem on Evans. If the mapping is injective why do we need to prove uniqueness again?

This is the theorem and its proof (From Evans L., Partial Differential Equations, p. $297-299$)

Theorem 1 (Lax-Milgram Theorem). Assume that$$B: H × H → \mathbb{R} \tag{i}$$ is a bilinear mapping, for which there exists constant $α, β > 0$ such that$$|B[u, v]| \le α \| u \| \| v \|$$ and $$\beta\|u\|^2\leq B[u,u]\ \,(u\in H).\tag{ii}$$ Finally, let $f:H\to\mathbb{R}$ be a bounded linear functional on $H$.

Then there exists a unique element $u\in H$ such that $$B[u,v]=\langle f,v\rangle\tag{1}$$ for all $v\in H$.

Proof.

1. For each fixed element $u\in H$, the mapping $v\mapsto B[u,v]$ is a bounded linear functional on $H$; whence the Riesz Representation Theorem (D.3) asserts the existence of a unique element $w\in H$ satisfying $$B[u,v]=(w,v)\ \, (v\in H).\tag{2}$$ Let us write $Au=w$ whenever $(2)$ holds; so that $$B[u,v]=(Au,v)\ \, (u,v\in H).\tag{3}$$

2. We first claim $A:H\to H$ is a bounded linear operator. Indeed if $\lambda_1,\lambda_2\in\mathbb{R}$ and $u_1,u_2\in H$, we see for each $v\in H$ that \begin{align}(A(\lambda_1u_1+\lambda_2u_2),v)&=B[\lambda_1u_1+\lambda_2u_2,v]\ \text{ by }(3)\\ &=\lambda_1B[u_1,v]+\lambda_2 B[u_2,v] \\ &=\lambda_1(Au_1,v)+\lambda_2(Au_2,v)\ \text{ by }(3)\text{ again} \\&=(\lambda_1 Au_1+\lambda_2 Au_2,v). \end{align} This equality obtains for each $v\in H$, and so $A$ is linear. Furthermore $$\|Au\|^2=(Au,Au)=B[u,Au]\leq\alpha\|u\|\,\|Au\|.$$ Consequently $\|Au\|\leq\alpha\|u\|$ for all $u\in H$, and so $A$ is bounded.

3. Next we assert $$\left\{\begin{array}{l}A\text{ is one-to-one, and}\\ R(A),\text{ the range of }A,\text{ is closed in }H. \end{array}\right.\tag{4}$$ To prove this, let us compare $$\beta\|u\|^2\leq B[u,u]=(Au,u)\leq\|Au\|\,\|u\|.$$ Hence $\beta\|u\|\leq\|Au\|$. This inequality easily implies $(4)$.

4. We demonstrate now $$R(A)=H.\tag{5}$$ For if not, then since $R(A)$ is closed, there would exist a nonzero element $w\in H$ with $w\in R(A)^{\bot}$. But this fact in turn implies the contradiction $\beta\|w\|^2\leq B[w,w]=(Aw,w)=0$.

5. Next, we observe once more from the Riesz Representation Theorem that $$\langle f,v\rangle =(w,v)\quad\text{for all }v\in H$$ for some element $w\in H$. We then utilize $(4)$ and $(5)$ to find $u\in H$ satisfying $Au=w$. Then $$B[u,v]=(Au,v)=(w,v)=\langle f,v\rangle\quad(v\in H),$$ and this is $(1)$.

6. Finally, we show there is at most one element $u\in H$ verifying $(1)$. For if both $B[u,v]=\langle f,v\rangle$ and $B[\tilde{u},v]=\langle f,v\rangle$, then $B[u-\tilde{u},v]=0\ (v\in H)$. We set $v=u-\tilde{u}$ to find $\beta\|u-\tilde{u}\|^2\leq B[u-\tilde{u},u-\tilde{u}]=0.$ $\tag*{$\square$}$

If we already know that $\langle f,v \rangle = (w,v)$ = $B[u,v] = (Au, v)$ and we know that $A$ is one-to-one, isn't that already a proof that there can be only one $u$ for which $$B(u,v) = (w,v) \tag{2}$$ holds?

Why do we need point 6 in the proof of the above theorem? Isn't that already explained?

• Just for clarification: (.,.) denotes the inner product in $L^2(U)$ and $<f,v> = \int_U f^0v + \sum_{i=1}^n f^i v_{x_i} dx$ – Max Herrmann Oct 19 '15 at 6:36
• @ConcordoCosta: Yes, the injectivity is already proven in 3. – gerw Oct 19 '15 at 6:57

Step 6 is indeed redundant. One can see that the argument used in Step 6 is essentially identical to that used for proving the 1-1 property of $A$.