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Mathematics LibreTexts

6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps

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Exercise 6.7.E.1

Discuss: In Definition 1, ¯G can equivalently be replaced by G=Gp(δ) (an open globe).

Exercise 6.7.E.2

Prove that if the set D is open (closed) in (S,ρ), then the map f:ST is open (closed, respectively) on D iff fD(f restricted to D) has this property as a map of D into f[D].
[Hint: Use Theorem 4 in Chapter 3, §12.]

Exercise 6.7.E.3

Complete the missing details in the proofs of Theorems 1-4.

Exercise 6.7.E.3

Verify footnotes 2 and 3.

Exercise 6.7.E.4

Show that a map f:EE may fail to be one-to-one on all of E even if f satisfies Theorem 1 near every pE. Nonetheless, show that this cannot occur if E=E=E1.
[Hints: For the first part, take E=C,f(x+iy)=ex(cosy+isiny). For the second, use Theorem 1 in Chapter 5, §2.]

Exercise 6.7.E.4

(i) For maps f:E1E1, prove that the existence of a bijective df(p;) is equivalent to f(p)0.
(ii) Let
f(x)=x+x2sin1x,f(0)=0.
Show that f(0)0, and fCD1 near any p0; yet f is not one-to-one near 0. What is wrong?

Exercise 6.7.E.5

Show that a map f:En(Cn)En(Cn),fCD1, may be bijective even if det[f(p)]=0 at some p, but then f1 cannot be differentiable at q=f(p).
[Hint: For the first clause, take f(x)=x3,p=0; for the second, note that if f1 is differentiable at q, then Note 2 in §4 implies that det [df(p;)]det[df1(q;)]=10, since ff1 is the identity map.]

Exercise 6.7.E.6

Prove Corollary 2 for the general case of complete E and E.
[Outline: Given a closed X¯G, take any convergent sequence {yn}f[X]. By Problem 8 in Chapter 4, §8, f1(yn)=xn is a Cauchy sequence in X (why?). By the completeness of E,(xX)xnx (Theorem 4 of Chapter 3, §16). Infer that lim so f[X] is closed.]

Exercise \PageIndex{7}

Prove that "the composite of two open (closed) maps is open (closed)." State the theorem precisely. Prove it also for the uniform Lipschitz property.

Exercise \PageIndex{8}

Prove in detail that f :(S, \rho) \rightarrow\left(T, \rho^{\prime}\right) is open on D \subseteq S iff f maps the interior of D into that of f[D]; that is, f\left[D^{0}\right] \subseteq(f[D])^{0}.

Exercise \PageIndex{9}

Verify by examples that f may be:
(i) closed but not open;
(ii) open but not closed.
[Hints: (i) Consider f= constant. (ii) Define f : E^{2} \rightarrow E^{1} by f(x, y)=x and let
D=\left\{(x, y) \in E^{2} | y=\frac{1}{x}, x>0\right\};
use Theorem 4(iii) in Chapter 3, §16 and continuity to show that D is closed in E^{2}, but f[D]=(0,+\infty) is not closed in E^{1}. However, f is open on all of E^{2} by Problem 8. (Verify!)]

Exercise \PageIndex{10}

Continuing Problem 9(ii), define f : E^{n} \rightarrow E^{1} (or C^{n} \rightarrow C) by f(\vec{x})= x_{k} for a fixed k \leq n (the "kth projection map"). Show that f is open, but not closed, on E^{n}\left(C^{n}\right).

Exercise \PageIndex{11}

(i) In Example (a), take (p, q)=(5,0) or (-5,0). Are the conditions of Theorem 4 satisfied? Do the conclusions hold?
(ii) Verify Example (b).

Exercise \PageIndex{12}

(i) Treating z as a function of x and y, given implicitly by
f(x, y, z)=z^{3}+x z^{2}-y z=0, \quad f : E^{3} \rightarrow E^{1},
discuss the choices of P and Q that satisfy Theorem 4. Find \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y}.
(ii) Do the same for f(x, y, z)=e^{x y z}-1=0.

Exercise \PageIndex{13}

Given f : E^{n}\left(C^{n}\right) \rightarrow E^{m}\left(C^{m}\right), n>m, prove that if f \in C D^{1} on a globe G, f cannot be one-to-one.
[Hint for f : E^{2} \rightarrow E^{1}: If, say, D_{1} f \neq 0 on G, \operatorname{set} F(x, y)=(f(x, y), y).]

Exercise \PageIndex{14}

Suppose that f satisfies Theorem 1 for every \vec{p} in an open set A \subseteq E^{\prime}, and is one-to-one on A (cf. Problem 4). Let g=f_{A}^{-1} (restrict f to A and take its inverse). Show that f and g are open and of class C D^{1} on A and f[A], respectively.

Exercise \PageIndex{15}

Given \vec{v} \in E and a scalar c \neq 0, define T_{\vec{v}} : E \rightarrow E ("translation by \vec{v}") and M_{c} : E \rightarrow E ("dilation by c"), by setting
T_{\vec{v}}(\vec{x})=\vec{x}+\vec{v} \text { and } M_{c}(\vec{x})=c \vec{x}.
Prove the following.
(i) T_{\vec{v}} and T_{\vec{v}}^{-1}\left(=T_{-\vec{v}}\right) are bijective, continuous, and "clopen" on E; so also are M_{c} and M_{c}^{-1}\left(=M_{1 / c}\right).
(ii) Similarly for the Lipschitz property on E.
(iii) If G=G_{\vec{q}}(\delta) \subset E, then T_{\vec{v}}[G]=G_{\vec{q}+\vec{v}}(\delta), and M_{c}[G]=G_{c \vec{q}}(|c \delta|).
(iv) If f : E^{\prime} \rightarrow E is linear, and \vec{v}=f(\vec{p}) for some \vec{p} \in E^{\prime}, then T_{\vec{v}} \circ f=f \circ T_{\vec{p}}^{\prime} and M_{c} \circ f=f \circ M_{c}^{\prime}, where T_{\vec{p}}^{\prime} and M_{c}^{\prime} are the corresponding maps on E^{\prime}. If, further, f is continuous at \vec{p}, it is continuous on all of E^{\prime}.
[Hint for (iv): Fix any \vec{x} \in E^{\prime}. Set \vec{v}=f(\vec{x}-\vec{p}), g=T_{\vec{v}} \circ f \circ T_{\vec{p}-\vec{x}}^{\prime}. Verify that g=f, T_{\vec{p}-\vec{x}}^{\prime}(\vec{x})=\vec{p}, and g is continuous at \vec{x}.]

Exercise \PageIndex{16}

Show that if f : E^{\prime} \rightarrow E is linear and if f\left[G^{*}\right] is open in E for some G^{*}=G_{\vec{p}}(\delta) \subseteq E^{\prime}, then
(i) f is open on all of E^{\prime};
(ii) f is onto E.
[Hints: (i) By Problem 8, it suffices to show that the set f[G] is open, for any globe G (why?). First take G=G_{\overrightarrow{0}}(\delta). Then use Problems 7 and 15(i)-(iv), with suitable \vec{v} and c.
(ii) To prove E=f\left[E^{\prime}\right], fix any \vec{y} \in E. As \(f=G_{\overrightarrow{0} (\delta\) is open, it contains a globe G^{\prime}=G_{\overrightarrow{0}}(r). For small c, c \vec{y} \in G^{\prime} \subseteq f\left[E^{\prime}\right]. Hence \vec{y} \in f\left[E^{\prime}\right] (Problem 10 in §2).]

Exercise \PageIndex{17}

Continuing Problem 16, show that if f is also one-to-one on G^{*}, then
f : E^{\prime} \stackrel{\longleftrightarrow}{\text { onto }} E,
f \in L\left(E^{\prime}, E\right), f^{-1} \in L\left(E, E^{\prime}\right), f is clopen on E^{\prime}, and f^{-1} is so on E.
[Hints: To prove that f is one-to-one on E^{\prime}, let f(\vec{x})=\vec{y} for some \vec{x}, \vec{x}^{\prime} \in E^{\prime}. Show that
(\exists c, \varepsilon>0) \quad c \vec{y} \in G_{\overrightarrow{0}}(\varepsilon) \subseteq f\left[G_{\overrightarrow{0}}(\delta)\right] \text { and } f(c \vec{x}+\vec{p})=f\left(c \vec{x}^{\prime}+\vec{p}\right) \in f\left[G_{\vec{p}}(\delta)\right]=f\left[G^{*}\right].
Deduce that c \vec{x}+\vec{p}=c \vec{x}^{\prime}+\vec{p} and \vec{x}=\vec{x}^{\prime}. Then use Problem 15(v) in Chapter 4, §2, and Note 1.]

Exercise \PageIndex{18}

A map
f :(S, \rho)\stackrel{\longleftrightarrow}{\text { onto }} (T, \rho^{\prime})
is said to be bicontinuous, or a homeomorphism, (from S onto T) iff both f and f^{-1} are continuous. Assuming this, prove the following.
(i) x_{n} \rightarrow p in S iff f\left(x_{n}\right) \rightarrow f(p) in T;
(ii) A is closed (open, compact, perfect) in S iff f[A] is so in T;
(iii) B=\overline{A} in S iff f[B]=\overline{f[A]} in T;
(iv) B=A^{0} in S iff f[B]=(f[A])^{0} in T;
(v) A is dense in B (i.e., A \subseteq B \subseteq \overline{A} \subseteq S) in (S, \rho) iff f[A] is dense in f[B] \subseteq\left(T, \rho^{\prime}\right).
[Hint: Use Theorem 1 of Chapter 4, §2, and Theorem 4 in Chapter 3, §16, for closed sets; see also Note 1.]

Exercise \PageIndex{19}

Given A, B \subseteq E, \vec{v} \in E and a scalar c, set
A+\vec{v}=\{\vec{x}+\vec{v} | \vec{x} \in A\} \text { and } c A=\{c \vec{x} | \vec{x} \in A\}.
Assuming c \neq 0, prove that
(i) A is closed (open, compact, perfect) in E iff c A+\vec{v} is;
(ii) B=\overline{A} iff c B+\vec{v}=\overline{c A+\vec{v}};
(iii) B=A^{0} iff c B+\vec{v}=(c A+\vec{v})^{0};
(iv) A is dense in B iff c A+\vec{v} is dense in c B+\vec{v}.
[Hint: Apply Problem 18 to the maps T_{\vec{v}} and M_{c} of Problem 15, noting that A+\vec{v}= T_{\vec{v}}[A] and c A=M_{c}[A].]

Exercise \PageIndex{20}

Prove Theorem 2, for a reduced \delta, assuming that only one of E^{\prime} and Eis E^{n}\left(C^{n}\right), and the other is just complete.
[Hint: If, say, E=E^{n}\left(C^{n}\right), then f[\overline{G}] is compact (being closed and bounded), and so is \overline{G}=f^{-1}[f[\overline{G}]]. (Why?) Thus the Lemma works out as before, i.e., f[G] \supseteq G_{\overline{q}}(\alpha).
Now use the continuity of f to obtain a globe G^{\prime}=G_{\vec{p}}\left(\delta^{\prime}\right) \subseteq G such that f\left[G^{\prime}\right] \subseteq G_{\vec{q}}(\alpha). Let g=f_{G}^{-1}, further restricted to G_{\vec{q}}(\alpha). Apply Problem 15(v) in Chapter 4, §2, to g, with S=G_{\vec{q}}(\alpha), T=E^{\prime}.]


6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.

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