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Let $$X = C^{k+2, \alpha}(S(T)),$$ $$Y = C^{k, \alpha}(S(T)),$$ where $S(T) = S^1 \times [0,T]$. Don't think of $T$ as fixed, but varying. So these Banach spaces contains functions with different time intervals. Suppose there is a map $F:X \to Y$ with $$F(u) = u_t - a(x,t,u,u_x,u_{xx})$$ where $a(x,t,z,p,q)$ is smooth in its arguments. We want to show that there is a unique $u^*$ such that $F(u^*) = 0.$ To do this, we can show that the derivative at $u$ $$DF(u)v = v_t - a_z(u)v - a_p(u)v_x - a_q(u)v_{xx}$$ is invertible (or bijective or linear isomorphism) at a particular function $u$. It is invertible, and we also know that the inverse mappings $DF(u)^{-1}$ are varying continuously and are uniformly bounded (for bounded $u$) regardless of the time interval in the domain.

Now can someone please explain these points I don't understand:

If there is a $u^0 \in X$ such that $F(u^0)$ is small, then the inverse function theorem implies that for all small $s \in Y$, there exists a unique $u$ such that $F(u) = s$, and $u$ depends continuously on $s$.

Is that right? By "small", I guess the author means close to zero. My understanding is, if $F(u^0)$ is in a neighbourhood of zero, for all functions $s$ in that same neighbourhood of $0$, we can find a $u$ in some neighbourhood of $u^0$ such that $F(u) = s$. Is that correct? Why does $u$ depend continuously on $s$?

Now if $u^0 = a(x,t,0,0,0)t \in X$, provided $T$ is small enough, $F(u^0)$ is as close to $0$ as required. This is the point when we take the time interval $[0,T]$ to be short.

It is true that $F(u^0) \to 0$ as $t \to 0$. But this is pointwise convergence, don't we need convergence in the $Y$ norm? Also, how can we be sure that 0 in fact lies in the neighbourhood of $Y$ that becomes invertible?

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What is the source of your quotes? My first reading was a bit hasty. There seems now to be a problem with the second statement. (The first statement is probably okay). –  Willie Wong Jul 18 '12 at 14:52
@WillieWong Thanks for your help. The source is "Curvature-Driven Flows: A variational approach" by Fred Almgren, Jean E. Taylor, and Lihe Wang. If you can't find it, here's the most relevant pages s14.postimage.org/9yhvzrs1d/image.jpg, s13.postimage.org/qv7c066vb/image.jpg and s17.postimage.org/tyhpehpb3/image.jpg. –  Court Jul 18 '12 at 15:32
@WillieWong Note that I changed the notation and some things are different since I am using this paper in conjunction with another paper ("The Curve Shortening Problem" by Xi-Ping Zhu and Chou). –  Court Jul 18 '12 at 15:33
Also, the paper is in the book "Selected Works of Frederick J Almgren, Jr." –  Court Jul 18 '12 at 15:34
Sorry I got the order wrong. It should be: s17.postimage.org/tyhpehpb3/image.jpg, s13.postimage.org/qv7c066vb/image.jpg, and s14.postimage.org/9yhvzrs1d/image.jpg, –  Court Jul 18 '12 at 15:39
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1 Answer

up vote 2 down vote accepted
  1. Your interpretation is not quite right. Inverse function theorem (under your assumptions) states that for $u^0$, there exists a neighborhood $N^0 \ni F(u^0)$ such that $F$ is invertible on $N^0$ with continuous inverse (in fact continuously differentiable). By the uniform bound on $DF^{-1}$ you can take the size of $N^0$ to be uniform: in particular there exists a constant $\delta$ such that $N^0 \supseteq F(u_0) + B_\delta$. So if $\|F(u_0)\| < \delta$, $0\in N^0$ and there exists $\epsilon$ such that $B_\epsilon \subseteq N^0$, meaning that for $s\in B_\epsilon \subseteq N^0$ we have a continuous map $F^{-1}: B_\epsilon \to X$.

  2. Yes, you need convergence in the $Y$ norm. Note that (writing $a(x,t) = a(x,t,0,0,0)$) $$F(ta(x,t)) = a(x,t) + ta_t(x,t) - a(x,t,ta, ta_x, ta_{xx})$$ The first and third terms combine to give (using the differentiability of $a$) a quantity that is $O(t)$. This shows that in particular that all $x$ derivatives of $F(ta(x,t))$ are $O(t)$ and hence can be made as small as you want by making $t$ small.

    There is, however, a problem with the $t$ derivatives, if you take $F(ta(x,t))_t |_{t = 0}$ you get $$ a_t(x,0) + a_t(x,0) - a_t(x,0) - a_u(x,0) a(x,0) - a_p(x,0) a_x(x,0) - a_q(x,0) a_{xx}(x,0) $$ which is not in general 0. So whichever source you are quoting from is incomplete in its argument.

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Thanks a lot. I'm afraid I have a few questions: 1) I guess you used some variant of the bounded inverse theorem to say that $DF(u)$ is uniformly bounded (since we only know that $DF^{-1}(u)$ is uniformly bounded). 2) what do you mean by take the size of $N^0$ to be uniform? Is there some theorem that says that uniformly bounded derivative gives such a property about taking $N^0$ to be uniform? –  Court Jul 18 '12 at 12:37
3) In the last sentence of your first point (1.), I should read that as "So if $\lVert F(u_0) \rVert < \delta, then 0 \in N^0$, ...". I'll need to think more about what you wrote but just checking whether that's right. 4) Sorry about not defining $G$, $G$ is actually the smooth function $a$ that I defined. By "$G(x,t,0,0,0)$ lies in bounded ball inside Y" do you not mean that "$\lVert G(x,t,0,0,0)\rVert_Y \leq \text{const}$? Sorry for so many questions. –  Court Jul 18 '12 at 12:37
(1) I meant $DF^{-1}$, not $DF$. Sorry about the typo. (3) Yes. (2) The meaning of uniform $N^0$ is after the colon in the same sentence. You need to use uniform boundedness of $DF^{-1}$ in addition to the fact that your $F$ is obviously continuously differentiable, and furtherore $DF$ is uniformly bounded on any bounded subset of $X$. These use the properties of $a$. With those bounds you can get an effective version of the Inverse Function Theorem. –  Willie Wong Jul 18 '12 at 14:26
Ah wait, sorry, I see where the confusion is with point 4. Let me edit the answer to fix that. –  Willie Wong Jul 18 '12 at 14:33
@TagWoh: Uniformity is the statement that the $\delta$ is a constant (in particular should not depend on $F(u^0)$). In the usual statement of IFT (which requires only strong differentiability at a single point) the $\delta$ is allowed to depend on everything. –  Willie Wong Jul 20 '12 at 15:16
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