# A “false” Lax-Milgram proof

A correct proof of the Law-Milgram Theorem using the Banach-Fixed Point Theorem could be found here on slide 10 onwards. I have another proof, which must be false, but I cannot see the fault...

Let $V$ be a vector space with inner product $\langle\cdot, \cdot \rangle$ and induced norm $\|u\| := \sqrt{\langle u,u \rangle}$ and $a : V\times V \to V$ a bounded bilinear form, i.e. there exists some $C >$ such that $|a(u,v)| \le C\|u\|\|v\|$.

For $v \in V$ define a linear map $(Av)(u) := a(v,u)$. Then by Riesz representation theorem there exists a unique $\tau Av$ such that for all $u \in V$ we have $$\langle \tau Av, u \rangle = (Av)(u)$$ and this correspondence is linear in $u$ and in $v$ and isometric. So we have

1) $\|\tau Av\| = \|Av\|$ because the ''Riesz correspondence'' is isometric,

2) $\|Av\| \le C\|v\|$ because $(Av)u = a(v,u) \le C\|v\|\|u\|$.

Then I want to show that there exists some $\varepsilon$ such that the map $$T_{\varepsilon}u := u - \varepsilon(\tau Au - f)$$ (where $f$ is the ''Riesz'' representant of some given linear functional $F : V\to V$) is a contraction mapping. For this \begin{align*} \| T_{\varepsilon}u - T_{\varepsilon}v \| & = \| (u-v) - \varepsilon( \tau Au - \tau Av ) \| \\ & = \| (u-v) - \varepsilon( \tau A(u-v) ) \| \\ & \le \| (u-v) \| + \varepsilon \|\tau A(u-v) \| \\ & = \| (u-v) \| + \varepsilon \|A(u-v)\| \\ & \le \| (u-v) \| + \varepsilon C\|u-v\| \\ & = (1+\varepsilon C)\|u-v\| \end{align*} so each $\varepsilon < 0$ would work, but this result is false (see for example the correct proof in the slides, and btw that $a$ must be coercive was not required in my arguments), but I cannot see what went wrong in the argumentation, I just used valid equations for the norms and so on? Can you help me finding my fault?

You assumed $\epsilon>0$ when you took $\epsilon$ outside of the norm. It should really be $|\epsilon|\,.$ This would give $\|Tu-Tv\|\le (1+|\epsilon|C)\|u-v\|\,.$