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 $v\in H$.


  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?

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

Evans might just think it is clearer for students to put in the extra step.


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