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$\newcommand{\INV}{{}^{-1}}$I'm reading an old script of a lecture long ago, and I need to understand the proof of the following version of the implicit function theorem, but calculus has admittedly always been a weak point.

Let $V_1,V_2,W$ be Banach spaces, and $\Omega\subset V_1\times V_2$ an open subset containing $(x_0,y_0)$. We also have a continuously differentiable map $$f:\Omega \to W$$ with $f(x_0,y_0) = 0$ and s.t. the derivative on the second component $$D_2f_{(x_0,y_0)}: y \mapsto Df _{(x_0,y_0)}(0,y)\in W$$ is homeomorphic onto its image.

Then there exists an open set $\Omega_1\times \Omega_2\subset \Omega$ around $(x_0,y_0)$ and a unique map $$g:\Omega_1\to \Omega_2$$ s.t $f(x,g(x))=0$ for all $x\in \Omega_1$.

PROOF

Now the proof is split into 3 steps.

  1. Let $L \equiv D_2f_{(x_0,y_0)}$, and define $$G(x,y)\equiv y - (L^{-1}\circ f)(x,y).$$ Clearly $G(x,y)=y$ iff $f(x,y)=0$.

To now invoke the Banach fixed point theorem, we basically need to check that $G(x,\cdot)$ maps a closed ball to itself, for all $x$ in some open set.

Now is where my confusion begins.

  1. We have $$ G(x,y_1) - G(x,y_2) = L\INV \left( L(y_1 - y_2) - f(x,y_1) + f(x,y_2)\right).$$ $f$ is differentiable and $L$ is continuous, so there exist $\delta_1,\eta>0$ s.t. $$ \tag{1} \lVert G(x,y_1) - G(x,y_2) \rVert \leq\frac{1}{2}\lVert y_1 - y_2\rVert $$ for all $x\in U(x_0,\delta_1)$, $y_1,y_2\in U(y_0,\eta)$ [open balls]. There also exists a $\delta_2>0$ s.t. for all $x\in U(x_0,\delta_2)$ it is true that $$ \tag{2}\lVert G(x,y_0)-G(x_0,y_0)\rVert\leq \frac{\eta}{2}. $$ Let $\delta \equiv \min\{\delta_1,\delta_2\}$. If now $y\in B(y_0,\eta)$ [closed ball], then \begin{align*} \lVert G(x,y)- y_0 \rVert &= \lVert G(x,y)- G(x_0,y_0) \rVert \\ &\leq \lVert G(x,y)- G(x,y_0) \rVert + \lVert G(x,y_0)- G(x_0,y_0) \rVert \\ &\leq \frac{1}{2}\lVert y-y_0\rVert + \frac{\eta}{2}\\ &\leq \eta \end{align*} for all $x\in B(x_0,\delta)$, which just means $G(x,y)\in B(y_0,\eta)$, so $G(x,\cdot)$ maps the closed ball of radius $\eta$ around $y_0$ to itself.

The rest is

  1. Apply Banach's fixed point theorem.

Questions

  1. In equation $(1)$, I assume the right hand side is the epsilon for which we have taken the deltas, coming from continuity in $(x_0,y_0)$, right? But then this raises the question: Why the $\leq$ sign instead of $<$?

  2. The same questions apply to equation $(2)$

  3. Then, after equation $(2)$ we even start using closed balls.

All of this boils down to using $\leq$ where $<$ should(?) have been used.

Or is this some kind of consequence of the Banach-ness of the involved spaces?

Thanks!


Edit:

Equation $(1)$ is true for all $y\in U(y_0,\eta).$ It is then not necessarily true for $y\in B(y_0,\eta),$ i.e. for $y\in\partial U(y_0,\eta).$ But in the last chain of inequalities (where we "prove" the assumptions for the FPT) we are assuming just that when going from the 2nd to 3rd line.

Even more, if we made the closed ball smaller by looking at $y\in B(y_0,\frac{\eta}{2})$, then the very last line would be $\leq \frac{3}{4}\eta$, which would most certainly not prove that $G(x,\cdot)$ maps the closed ball $B(y_0,\frac{\eta}{2})$ to itself.

Can we just choose $\delta_2$ s.t. in equation $(2)$ it says $\leq\frac{\eta}{4}$? There's no harm in doing that I suppose.

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    $\begingroup$ $a<b$ implies $a≤b$. A closed ball is used because of the fixed point theorem's assumptions, but you can check that nothing is wrong. $\endgroup$ – Calvin Khor Jan 24 '18 at 15:56
  • $\begingroup$ D'oh. But if, for example, $(2)$ is true for $x\in U(x_0,\delta_2)$, then it need not be true for $x\in B(x_0,\delta_2)$, right? I know that the closed ball around $y_0$ is important for the fixed point theorem, but I don't see why we use the closed ball around $x_0$ as well. Well, I guess there's a high chance it's a typo. $\endgroup$ – Jo Be Jan 24 '18 at 16:02
  • $\begingroup$ In addition when something is true on an open ball, you can make the radius a tiny bit smaller and use a closed ball instead. There is some flexibility in some definitions of analysis which is why you will see people write slightly different things than what you expect. In a sense, the important thing is somehow the picture, not the precise definition. $\endgroup$ – Calvin Khor Jan 24 '18 at 16:11
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    $\begingroup$ Actually, reading your proof again you are right, but by choosing $\delta/2$ instead the problem is fixed $\endgroup$ – Calvin Khor Jan 24 '18 at 16:18
  • $\begingroup$ Sorry for going at this again. Equation $(1)$ is true for all $y\in U(y_0,\eta)$. It is then not necessarily true for $y\in B(y_0,\eta)$. But in the last chain of inequalities (where we "prove" the assumptions for the FPT) we assume that it is, going from the 2nd to 3rd line. Even more, if we were looking at $y\in B(y_0,\frac{\eta}{2})$ (where the first equation holds), then the very last line would be $\leq \frac{3}{4}\eta$, which would most certainly not prove that $G(x,\cdot)$ maps $B(y_0,\frac{\eta}{2})$ to itself. $\endgroup$ – Jo Be Jan 25 '18 at 10:11
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In the chat, doubt was raised on the following part of the proof:

  1. We have $$ G(x,y_1) - G(x,y_2) = L\INV \left( L(y_1 - y_2) - f(x,y_1) + f(x,y_2)\right).$$ $f$ is differentiable and $L$ is continuous, so there exist $\delta_1,\eta>0$ s.t. $$ \tag{1} \lVert G(x,y_1) - G(x,y_2) \rVert \leq\frac{1}{2}\lVert y_1 - y_2\rVert $$ for all $x\in U(x_0,\delta_1)$, $y_1,y_2\in U(y_0,\eta)$ [open balls].

Here is the proof of this inequality. As $L$ is a continuous linear operator that is homeormorphic onto the image (in particular bijective) the Bounded Inverse Theorem guarantees that $L^{-1}$ is also continuous.

Now also \begin{align}f(x,y_2) - f(x,y_1) &= f(x,y_2) - f(x,y_0) + f(x,y_0) - f(x,y_1) \\ &= D_2 f(x,y_0)(0,y_2 - y_1) + o(y_2-y_0) + o(y_1-y_0) \end{align} By the fact that $f$ is continuously differentiable, we can make $L-D_2 f(x,y_0) $ arbitrarily small in operator norm. This is where the 1/2 is. Choose $\|x-x_0\|≤\delta_1$ so that the $o(\cdot)$ errors are independent of $x$ and $\|L-D_2 f(x,y_0)\|<\frac1{100\|L^{-1}\|}$, and also choose $y_1,y_2$ to lie in a radius $\eta\ll 1$ ball around $y_0$ so that the $o(\cdot)$ errors are smaller than $\frac1{100}\|y_1-y_2\|$. Then \begin{align} \| G(x,y_1) - G(x,y_2)\| &≤\|L^{-1}\|\|L-D_2f(x,y_0)\|\|y_1-y_2\| + o(y_2-y_0) + o(y_1-y_0) \\ &≤ \frac{1}{2} \|y_1-y_2\| \end{align}

The remainder of the proof I think is more straightforward, especially after the comments. From the proof of the inequality though, it is much simpler to completely avoid talking about small open balls, when we could have chosen small enough closed balls for the results to hold.

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  • $\begingroup$ Is $f(x,y_2) - f(x,y_0) + f(x,y_0) - f(x,y_1) = D_2 f(x,y_0)(0,y_2 - y_1)$ $+ o(y_2-y_0) + o(y_1-y_0)$ just an application of the definition of $D_2f$? $\endgroup$ – Jo Be Jan 25 '18 at 20:32
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    $\begingroup$ Yeah, also I didn't use a separate line for linearity of $D_2f(x,y_0)$. $\endgroup$ – Calvin Khor Jan 25 '18 at 20:36
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    $\begingroup$ @JoBe [I was even trying to be explicit, but] to neglect the $o(\cdot)$ errors you need to actually make $\|L-D_2f(x,y_0)\|$ even smaller. $\endgroup$ – Calvin Khor Jan 25 '18 at 20:47
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    $\begingroup$ @JoBe yes this is my previous comment. I wasn't careful enough, I can change the answer to reflect this. $\endgroup$ – Calvin Khor Jan 26 '18 at 9:31
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    $\begingroup$ @JoBe you could have fixed it any number of ways, for instance if you kept the $\frac1{2\|L^{-1}\|}$ bound then if you choose $\xi>0$ so that $$ \|L-D_2f(x,y_0)\| + \xi = \frac1{2\|L^{-1}\|} $$ it suffices to have $$\|o(\cdot)\| ≤ \xi \|L^{-1}\|\|y_1-y_2\|$$ I just chose to go with the realisation that the $1/2$ can be made arbitrarily smaller $\endgroup$ – Calvin Khor Jan 26 '18 at 9:47

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