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Dec 5, 2016 at 22:39 history edited Delio Mugnolo CC BY-SA 3.0
corrected spelling
Oct 10, 2012 at 20:04 comment added Delio Mugnolo (and all $u\in V$, of course).
Oct 10, 2012 at 20:04 comment added Delio Mugnolo Take the quadratic form associated with the operator $A_t$, say $a_t$. This is defined on a Hilbert space, say $V_t$, but as I wrote above in Lions' scenario one is assuming that $V_t\equiv V$. For a certain fixed $t$, coercivity of $a_t$ means that $$a_t(u)\ge \alpha_t \|u\|_V$$ for some $\alpha_t>0$ and all $u\in V$. Equi-coercivity of $(a_t)$ means that the same estimate holds for all $t$ for some constant $\alpha$ not depending on $t$, i.e., $$a_t(u)\ge \alpha \|u\|_V$$ for some $\alpha>0$ and all $t$.
Oct 10, 2012 at 19:52 comment added André Henriques May I ask: what does "equi-coercive" mean?
Oct 10, 2012 at 19:44 history edited Delio Mugnolo CC BY-SA 3.0
added 270 characters in body
Oct 10, 2012 at 19:38 history answered Delio Mugnolo CC BY-SA 3.0