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I'm studying the inf-sup conditions for saddle point problems. I'm referring to the usual one $$\begin{cases}Au + B^t p = f \\Bu=g \end{cases}$$ In the book I'm using (Ern - Guermond: Theory and practice of FEM) they say (pg. 100)

The problem is well posed iff

  • $$ \exists \alpha >0: \inf_{u \in \ker(B)} \sup_{v \in \ker(B)} \frac{a(u,v)}{||u||_X ||v||_X} \geq \alpha$$
  • $$\forall v \in \ker(B), \forall u \in \ker(B), a(u,v)=0 \implies v=0$$ [... and then we have also the inf-sup on $B$, which is not a problem for me right now]

I understand how they derive them, but then, looking at my lecture notes, my professor replaced the second condition with another inf-sup

$$ \exists \alpha >0: \inf_{v \in \ker(B)} \sup_{u \in \ker(B)} \frac{a(u,v)}{||u||_X ||v||_X} \geq \alpha$$

I've searched in lots of books and lecuture notes, but I still cannot understand why this one above is equivalent to the second condition. In particular, it's clear to me that the last "inf-sup" must be strictly greater than $0$, but I don't see why the constant $\alpha$ must be the same.

What am I missing?

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  • $\begingroup$ I don't think they're equivalent (in infinite dimensions), although the latter implies the former -- it's a sort of "quantitative injectivity". To see this, you can rearrange (estimating for fixed $u$ the $\inf$ over all $v$ by an arbitrary one and using the definition of the operator norm) to $\|A^*v\|\geq \alpha\|v\|$, which clearly implies the desired injectivity. $\endgroup$ – Christian Clason Jul 19 at 18:43
  • $\begingroup$ Just to be sure I understood your comment properly: if I rewrite the latter as $$ (\star) \inf_{v \in X} \frac{||A^* v||_{\ker(B)'}}{||v||} \geq \alpha > 0 $$ then $\ker(A^*)$ must be $\{0\}$ only, because $\star$ is $0$ only if $\ker(A^*)$ is not trivial. Indeed $\ker{A^*} = \{0\}$ is precisely the second condition of theorem. Is this what you had in mind? @ChristianClason $\endgroup$ – bobinthebox Jul 19 at 19:25
  • $\begingroup$ close enough -- I meant that (estimating the $\inf$ by some $w$ from above to make it clearer) this inequality shows that $\|A^*w\|=0$ implies that $\|w\|=0$, which is exactly injectivity. $\endgroup$ – Christian Clason Jul 19 at 19:28
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    $\begingroup$ Uh yes, because if you rearrange the inf-sup with $||A^* w|| \geq \alpha ||w||$ then indeed if $Av = Aw$ you got $0=||A(v-w)|| \geq \alpha ||v-w||$ which clearly implies $v = w$. @ChristianClason I think I got it now, right? $\endgroup$ – bobinthebox Jul 19 at 19:38
  • $\begingroup$ Yep, that's a perfectly valid way of putting it! $\endgroup$ – Christian Clason Jul 19 at 19:40

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