17.4 Implicit function theorem [2D4]

We will use the Implicit Function Theorem, in the multivariable version (Theorem 7.7.4 in [ 2 ] ). We recall it here for convenience, with some small changes in notations.

Theorem 399 Implicit function theorem in \(ℝ^ n\)

[1GD] Let \(f:A⊆ℝ^ n→ℝ\) be continuous, with \(A\) open, and let \(\overline x=(\overline x',\overline x_ n)∈ A\) be such that \(∂_{x_ n} f\) exists in a neighborhood of \(\overline x\), is continuous in \(\overline x\) and \(∂_{x_ n}f(\overline x) ≠0\). Define \(\overline a=f(\overline x)\).

There is then a ”cylindrical” neighborhood \(U\) of \(\overline x\)

\[ U=U'× J \]

where

\[ U'=B(\overline x',𝛼) \]

is the open ball in \(ℝ^{n-1}\) centered in \(\overline x'\) of radius \(𝛼{\gt}0\), and

\[ J= (\overline x_ n-𝛽,\overline x_ n+𝛽) \]

with \(𝛽{\gt}0\). Inside this neighborhood \(U∩ f^{-1}(\{ \overline a\} )\) coincides with the graph \(x_ n=g(x')\), with \(g:U'\to J\) continuous.

This means that, for every \(x=(x',x_ n)∈ U\), \(f(x)=\overline a\) if and only if \(x_ n=g(x')\).

Moreover, if \(f\) is of class \(C^ k\) on \(A\) for some \(k∈ℕ^*\), then \(g\) is of class \(C^ k\) on \(U'\) and

\begin{equation} \label{der_ g_ 2} \def\de {∂} \frac{∂g}{∂x_ i}(x')=-\frac{\displaystyle \frac{∂f}{∂x_ i}\big(x',g(x'))}{\displaystyle \frac{∂f}{∂x_ n}\big(x',g(x'))} \qquad \forall x'∈ U', \forall i, 1≤ i≤ n-1\quad . \end{equation}
400

E400

[1GF] Consider the following \(C^∞\) function of 2 variables

\[ {f}(x,y)= x^ 3+y^ 4-1 \quad . \]

Check that \(\{ f=0\} =\{ (x,y)∈ℝ^ 2: f(x,y)=0\} \) is not empty; then, for each point of the plane where \(f\) vanishes, discuss whether the implicit function theorem can be applied, and therefore if the set \(\{ f=0\} \) is locally graph of a \(C^∞\) function. Also study the set \(\{ f=0\} \): is it compact? How many connected components are there?

(Please note what is shown in 4).

Hidden solution: [UNACCESSIBLE UUID ’1GG’]

E400

[1GJ] Repeat the study of the previous exercise 1 for the function

\[ f(x,y)= \sin (x+y)+x^ 2\quad . \]

Hidden solution: [UNACCESSIBLE UUID ’1GK’][UNACCESSIBLE UUID ’1GN’]

E400

[1GP]Note:Exercise 2, Written exam, June 30th 2017.Repeat the study of the previous exercise for the function

\[ f(x,y)= 1 + 4x + e^ x y + y^ 4\quad . \]

Show that the zero set is not compact.

E400

[1GQ]Let \(A⊂ ℝ^ 3\) be an open set and suppose that \(f,g:A→ℝ\) is differentiable, and such that in \(p_ 0=(x_ 0,y_ 0,z_ 0)∈ A\) we have that \(∇ f(p_ 0),∇ g(p_ 0) \) are linearly independent and \(f(p_ 0)=g(p_ 0)=0\): show that the set \(E=\{ f=0,g=0\} \) is a curve in a neighborhood of \(p_ 0\).

(Hint: consider that the vector product \(w=∇ f(p_ 0)× ∇ g(p_ 0)\) is nonzero if and only if the vectors are linearly independent — in fact it is formed by the determinants of the minors of the Jacobian matrix. Assuming without loss of generality that \(w_ 3≠ 0\), show that \(E\) is locally the graph of a function \((x,y)=𝛾(z)\).)

Hidden solution: [UNACCESSIBLE UUID ’1GR’]

E400

[1GS] Note:Written exam, July 4th 2018.The figure 4 shows the set \(E=\big\{ (x,y): ye^ x+xe^ y=1\big\} \).

Properly prove the following properties:

  1. at every point \((x_ 0,y_ 0)∈ E\) the assumptions of the implicit function theorem are satisfied;

  2. \(E∩\big\{ (x,y): x{\gt}0\big\} \) coincides with the graph, in the form \(y=f(x)\), of a single function \(f\) defined on \((0,+∞)\);

  3. \(E\) is connected;

  4. \(\lim _{x→+∞} f(x)=0\).

  5. Show (at least intuitively) that \(x_ 0{\gt}0\) exists with the property that \(f\) is decreasing for \(0{\lt}x{\lt}x_ 0\), increasing for \(x{\gt}x_ 0\).

\includegraphics[width=6cm, height=6cm,keepaspectratio]{UUID/1/G/T/blob_zxx}
Figure 4 Figure for exercise 5.

Hidden solution: [UNACCESSIBLE UUID ’1GV’]

E400

[1GW] Let \(E\) be the set of horizontal lines

\[ E = \{ (x,0):x∈ℝ\} ∪ ⋃_{n=1}^∞ \{ (x,1/n):x∈ℝ\} \quad . \]

Find a function \(f:ℝ^ 2→ℝ\), \(f=f(x,y)\) class \(C^ 1\) such that \(E=\{ (x,y) : f(x,y)=0 \} \).

Prove that necessarily \(∂_{y} f(0,0)=0\).

Set \((\overline x,\overline y)=(0,0)\). Note that there is a function \(g:ℝ→ℝ\) such that \(g(0)=0\) and \(f(x,g(x))=0\)! In fact, the function \(g≡ 0\) is the only function with such characteristics. Thus part of the thesis in the implicit function theorem is satisfied.

So explain precisely why the thesis of the implicit function theorem is not satisfied.

Extensions

Now let’s see some variations of the ”standard” theorem.

E400

[1GX]Prerequisites:358.

We work in the hypotheses of the theorem 399. Show that, if \(f(⋅,y)\) is Lipschitz of constant \(L\) for every fixed \(y\), i.e.

\[ | f(x_ 1',y)- f(x_ 2',y)|≤ L |x_ 1'-x_ 2'|~ ~ ∀ x_ 1',x_ 2∈ U',y∈ J \]

(and \(L{\gt}0\) does not depend on \(x_ 1',x_ 2',y\)), then \(g\) is Lipschitz of constant \(L'\). What is the relationship between the constants \(L\) and \(L'\)?

Similarly if \(f\) is Hölderian.

Hidden solution: [UNACCESSIBLE UUID ’1GY’]

E400

[1GZ] In the same assumptions as the previous theorem 399, show that there exist \(\varepsilon {\gt}0\) and a continuous function \(\tilde g:V→ ℝ\) where \(I=(\overline a-\varepsilon ,\overline a+\varepsilon )\) and \(V=U'× I\) is open in \(ℝ^ n\), such that

\begin{equation} ∀ (x',a)∈ V \quad ,\quad (x',\tilde g(x',a))∈ U \quad \text{e} \quad f\big(x',\tilde g(x',a)\big)=a \quad .\label{eq:tilde_ g_ teor_ funz_ inv} \end{equation}
401

Vice versa if \(x∈ U\) and \(a=f(x)\) and \(a∈ I\) then \(x_ n=\tilde g(x',a)\).

Note that the previous relation means that, for each fixed \(x'∈ U'\), the function \(\tilde g(x',⋅)\) is the inverse of the function \(f(x',⋅)\) (when defined on appropriate open intervals).

So, moreover, the function \(\tilde g\) is always differentiable with respect to \(a\), and the partial derivative is

\[ \frac{\partial ~ }{\partial {a}} \tilde g(x',a)=\frac{1}{\frac{\partial ~ }{\partial {x_ n}} f(x',\tilde g(x',a))}\quad . \]

The other derivatives instead (obviously) are as in the theorem 399.

The regularity of \(\tilde g\) is the same as \(g\): if \(f\) is Lipschitz then \(\tilde g\) is Lipschitz; if \(f∈ C^ k(U)\) then \(\tilde g∈ C^ k(V)\).

Hidden solution: [UNACCESSIBLE UUID ’1H0’]

E400

[1H1] In the same hypotheses of the exercise 2, we also assume that \(f∈ C^ 1(A)\).

  • We decompose \(y=(y',y_ n),∈ℝ^ n\) as we did for \(x\). We define the function \(G:V→ ℝ^ n\) as \(G(y)=(y',\tilde g(y))\). Let \(W=G(V)\) be the image of \(V\), show that \(W⊆ U\) and that \(W\) is open.

  • Show that is \(G:V→ W\) is a diffeomorphism; and that its inverse is the map \(F(x)=(x',f(x))\).

  • Let’s define \(\tilde f = f ◦ G\). Show that \(\tilde f(x)=x_ n\).

(This exercise will be used, together with 6, to address constrained problems, in Section 17.5). Hidden solution: [UNACCESSIBLE UUID ’1H2’]

E400

[1H3] Prerequisites: 1, 1, 2, 2, 6, 399, 4 and 3.
Difficulty:**.

For this exercise we need definitions and results presented in the Chapter 21.

Let \(r≥ 1\) integer, or \(r=∞\). Let \(F:ℝ^ 2→ℝ\) of class \(C^ r\), and such that \(∇ F≠ 0\) at every point \(F=0\).

We know, from 1, that \(\{ F=0\} \) is the disjoint union of connected components, and from 1 that every connected component is a closed.

Show that, for every connected component \(K\), there is an open set \(A⊇ K\) such that \(K=A∩ \{ F=0\} \), and that therefore there are at most countably many connected components.

Show that each connected component is the support of a simple immersed curve of class \(C^ r\), of one of the following two types:

  • the curve is closed, or

  • The curve \(𝛾:ℝ→ℝ^ 2\) is not closed and is unbounded (i.e. \(\lim _{t→ ±∞}|𝛾(t)|=∞\)).

The first case occurs if and only if the connected component is a compact.

Hidden solution: [UNACCESSIBLE UUID ’1H4’][UNACCESSIBLE UUID ’1H5’]Please see PDF