Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $F:X \to Y$ is a map from Banach spaces $X=\widetilde{C}^{k+2, \alpha}(S)$ to $Y = \widetilde{C}^{k, \alpha}(S)$ where $S = I \times [0,T].$

Suppose the derivative $DF(u):X \to Y$ exists and is continuous.

(1) Am I right that the statement

$DF[u]^{-1}$ are uniformly bounded for bounded $u$

means $$\lVert DF[u]^{-1}\rVert \leq M$$ holds for all bounded (in what?) $u$ and $M$ is a constant not depending on $u$?

(2) If I write $DF[u]h = f$, and get a bound $$\lVert h \rVert \leq C\lVert h_0\rVert + C\lVert f \rVert$$ where $h_0 = h(\cdot, 0)$, then how does this show that $DF[u]^{-1}$ is uniformly bounded? (If the constant doesn't depend on $u$).

(3) Willie Wong said that if the $DF[u]^{-1}$ are uniformly bounded then applying the inverse function theorem to F, the size of the neighbourhood of $F(u^0)$ that is invertible doesn't depend on $F(u^0)$. Can someone give me a reference for this fact?

Thanks.

(These questions stem from Inverse function theorem in Banach space to prove short time existence of PDE (explanation of statements), and I didn't want to keep asking questions there!)

share|improve this question
    
How do you define the norm of $A^{-1}$ (assuming $A$ bijective)? –  Davide Giraudo Jul 23 '12 at 11:04
    
@DavideGiraudo Why not just $\sup_{\lVert T_x \rVert_{\mathcal{L}(X,Y) = 1}}\lVert A^{-1}(T_x) \rVert_{X}$ where $Ax = T_x$. So it's just $\sup_{\lVert Ax \rVert_{\mathcal{L}(X,Y) = 1}}\lVert x \rVert_X$? –  TagWoh Jul 23 '12 at 11:10
4  
This is very vague; it is like asking, "$x$ is a real number, what techniques can I use to show $x$ is positive"? With so little structure, there are too many techniques to list, and probably most of them would not be useful to you. You should probably say more about the specific problem you are studying. –  Nate Eldredge Jul 23 '12 at 12:10
1  
Not "if $DF^{-1}$ is bounded..."; it is "if additionally $DF^{-1}$ is bounded." Typically a sufficient assumption is that $DF$ is Lipschitz continuous. This would allow an effective proof via either Banach fixed point theorem or via Newton-Kantorovich. (My comment here was incomplete.) –  Willie Wong Jul 26 '12 at 13:44
1  
All the proofs that I know (could be just my own ignorance) of this fact requires using something which states that $\exists \delta > 0$ such that for all $u,v,w\in X$ with $\|v - u\| + \|w - u\| < \delta$, $$ \| (v-w) - DF^{-1}[u] (F(v) - F(w)) \| \leq \frac12 \|v-w\| $$ This suggests that we can probably weaken the $C^{1,1}$ condition on $F$ to something something similar to uniform continuity of $DF$ and still have the proof go through. Which also suggests that the additional regularity properties may be unnecessary if you restrict to $K\subset X$ compact. –  Willie Wong Jul 26 '12 at 13:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.