I'm studing the Brezzi Theorem (also called Babuška-Brezzi or Ladyzhenskaya-Babuška-Brezzi Theorem).

The Theorem 1 (on that link) ensures the continuous dependence of the solution, that is

$$\|(u,\lambda)\|_{X\times M}\leq C\{\|f\|_{X^*}+\|g\|_{M^*}\}.$$

With this expresion and recalling that

$$\|f\|_{X^*}=\sup_{v\in X}\dfrac{\langle f,v\rangle}{\|v\|_X}\quad\text{ and }\quad\|g\|_{M^*}=\sup_{\mu\in M}\dfrac{\langle g,\mu\rangle}{\|\mu\|_M}$$

we obtain that

$$\|(u,\lambda)\|_{X\times M}\leq C\left\{\sup_{v\in X}\dfrac{\langle f,v\rangle}{\|v\|_X}+\sup_{\mu\in M}\dfrac{\langle g,\mu\rangle}{\|\mu\|_M}\right\}$$

$$=C\left\{\sup_{v\in X}\dfrac{a(u,v)+b(v,\lambda)}{\|v\|_X}+\sup_{\mu\in M}\dfrac{b({\color{red}u},\mu)}{\|\mu\|_M}\right\}$$

My question is, it is possible to prove the following?

$$\boxed{\|(u,\lambda)\|_{X\times M}\leq C\left\{\sup_{(v,\mu)\in X\times M}\dfrac{a(u,v)+b(v,\lambda)+b({\color{red}u},\mu)}{\|(v,\mu)\|_{X\times M}}\right\}}$$

This is something like a global inf-sup condition.

  • 1
    $\begingroup$ The bound you seek seems a little unusual, but if you are interested in an overall bound on the inverse of the block operator $\begin{bmatrix}A & B^* \\ B & 0\end{bmatrix}$, where $A,B$ are the linear operators associated with the bilinear forms $a,b$, respectively, then I recommend you look at the paper 2013 paper of Krendl Et. Al. "Stability estimates and structural properties of saddle point systems", arxiv.org/pdf/1202.3330.pdf, which provides the sharpest known bounds in this regard and is quite readable. $\endgroup$
    – Nick Alger
    Dec 20, 2016 at 3:32
  • $\begingroup$ Thanks @NickAlger. Note that in that paper also says that the expresion is "well-known", but I can't see how all world can obtain it. I think that is not hard to prove, and for that reason nobody explain its prove, but I can't see it :( $\endgroup$
    – yemino
    Feb 18, 2017 at 1:14


You must log in to answer this question.