In The implicit function theorem written by Krantz & Parks, it's said that the implicit function theorem implies the following existence theorem of ODE:

Theorem 4.1.1 If $F(t,x)$, $(t,x)\in\mathbb R\times\mathbb R^N$, is continuous in the $(N+1)$-dimensional region $(t_0-a,t_0+a)\times B(x_0,r)$, then there exists a solution $x(t)$ of $$\frac{dx}{dt}=F(t,x),\qquad x(t_0)=x_0$$ defined over an interval $(t_0-h,t_0+h)$.

It's Peano existence theorem, I think. However, there seems a gap in their proof. WLOG, suppose $t_0=0$. They constructed $\mathcal H\colon[0,1]\times\mathcal B_1\to\mathcal B_0\times\mathbb R$, where $\mathcal B_0$ is the space of bounded continuous $\mathbb R^N$-valued functions on $(-a,a)$ normed canonically, and $\mathcal B_1$ is the space of bounded continuously differentiable $\mathbb R^N$-valued functions on $(-a,a)$ that also have a bounded derivative, normed canonically by $\sup\lvert f\rvert+\sup\lvert\dot f\rvert$, as follows: $$\mathcal H[\alpha,X(\tau)]=[X'(\tau)-\alpha F(\alpha\tau,X(\tau)),X(0)-x_0]$$. Note that $\mathcal H[0,x_0]=[0,0]$, where $x_0$ on the left side denotes the constant function. Then they claim that the existence theorem follows from the implicit function theorem. However, under the only condition that $F$ is continuous, there's no evidence that $\mathcal H$ is partially differentiable with respect to $X$ for $\alpha\in(0,x_0)$.

Can we fix the preceding proof in some extent?

PS: I posted the question not only because I want to comprehend such a proof, but also want to understand the relation between ODE and implicit functions. It seems certain that such a proof cannot be alright, since the canonical implicit function theorem is also a uniqueness theorem, which implies the local uniqueness of a solution of ODE. However, I want to know how to fix it. I doubt it might rely on a more general implicit function theorem.

  • $\begingroup$ what had you done in your own? $\endgroup$ – janmarqz Jan 17 '14 at 23:42
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    $\begingroup$ @janmarqz Well, I don't know how to answer your question. I did know Peano's existence theorem, and I did see a canonical proof of that, say, by Tonelli, through Arzela-Ascoli theorem, though I'm certainly unable to prove it on my own. What I want, is to explain the preceding proof, or fix the gap of that proof. $\endgroup$ – Yai0Phah Jan 19 '14 at 15:39

Observe that we have Fréchet-differentiability in the second argument at the point where you want your implicit function. Indeed, the function $T: Y \mapsto (Y', Y(0))$ is linear, and we have $$ \lim_{Y \to 0} \frac{\mathcal H(0, X + Y) - (\mathcal H(0, X) + T(Y))}{\|Y\|_1} = \lim_{Y \to 0} \frac{(0, 0)}{\|Y\|_1} = 0, $$ which is why $d_2 \mathcal H(0, x_0)$ exists and is equal to $T$. Note that $T$ is invertible, since $Y(0)$ determines the integration constant. Furthermore, we can obtain the following modified version of theorem 3.4.10:

Let $X, Y, Z$ be Banach spaces. Let $U \times V \subseteq X \times Y$, $g : U \times V \to Z$ and let $(x, y) \in U \times V$ such that $d_2(x, \cdot)$ exists and $G$ is continuous and $G(x, y) = 0$, and further such that there exists $W \subseteq V$ such that $y \in W$ and $G(z, w) \to G(x, w)$ as $z \to x$ uniformly for $w \in W$. Then there exist $M \subseteq U$ and $N \subseteq W$ such that for each $\xi \in X$, there exists a unique $\eta \in M$ such that $G(\xi, \eta) = 0$ and the thereby defined function is continuous.

In the proof, we replace the mean value theorem in part by the uniform convergence from the assumption, where we choose $N$ small enough such that $\|d_2 L(x, \cdot)\| < 1/4$.

From this theorem, we obtain Peano's existence theorem as in the given proof.

Note also that uniqueness does not follow, since the IFT only gives uniqueness locally around the constant function.

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    $\begingroup$ In order to apply implicit function theorem, one should assume that $\mathcal H$ is Fréchet-differentiable in a neighborhood of the point, so you cannot only check the differentiability of $\mathcal H(0,\bullet)$, but also that of $\mathcal H(\alpha,\bullet)$ for $\alpha>0$, according to Thm 3.4.10. $\endgroup$ – Yai0Phah Sep 26 '15 at 9:16
  • $\begingroup$ I'm currently working on fixing this. $\endgroup$ – Cloudscape Sep 26 '15 at 11:04
  • $\begingroup$ @FrankScience: I hope it's correct now, I'm sorry. This mistake occured because of my non-careful reading. $\endgroup$ – Cloudscape Sep 26 '15 at 12:25
  • $\begingroup$ I will probably check it at weekends but conceptually I don't think it works. Essentially, implicit function theorem is a theorem about solving non-linear equations, and the canonical proof furnishes an iteration process, so you can translate the proof to an iterative proof if you want, and I don't think such a proof exists for Peano's theorem. I think the compactness argument may be essential. $\endgroup$ – Yai0Phah Sep 29 '15 at 6:57
  • $\begingroup$ You don't need to check; I spotted another mistake. I think though that I have a way to prove this if $F$ is Lipschitz at zero, and that I can prove that the IFT is not applicable. Today I'm quite busy, hence I will edit my answer in the evening. $\endgroup$ – Cloudscape Sep 29 '15 at 10:17

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