Problem 38 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

Problems. 2-36.* Let \(A \subset \mathbf{R}^{n}\) be an open set and \(f: A \rightarrow \mathbf{R}^{n}\) a continuously differentiable 1-1 function such that det \(f^{\prime}(x) \neq 0\) for all \(x\). Show that \(f(A)\) is an open set and \(f^{-1}: f(A) \rightarrow A\) is differentiable. Show also that \(f(B)\) is open for any open set \(B \subset A\). 2-37. (a) Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be a continuously differentiable function. Show that \(f\) is not 1-1. Hint: If, for example, \(D_{1} f(x, y) \neq 0\) for all \((x, y)\) in some open set \(A\), consider \(g: A \rightarrow \mathbf{R}^{2}\) defined by \(g(x, y)=\) \((f(x, y), y)\) (b) Generalize this result to the case of a continuously differentiable function \(f: \mathbf{R}^{n} \rightarrow \mathbf{K}^{m}\) with \(m

Step-by-Step Solution

Answer

f(A) and f(B) open, f^-1 differentiable (f: A->ℝ°). 2-37a,b: f non 1-1 using rank generalization. Det non-zero but f not 1-1 in 2-38b.

Step 1 - Understanding the Exercise

The exercise involves proving different properties of continuously differentiable functions and involves concepts like open sets, differentiability, and the determinant of the derivative (Jacobian matrix).

Step 2 - Exercise 2-36

Given: A continuously differentiable 1-1 function f: A → ℝ^n where A is an open set subset of ℝ^n and det f'(x) ≠ 0 for all x ∈ A. To show: f(A) is an open set and f^{-1}: f(A) → A is differentiable.

Step 3: Substep 2.1 - Using Inverse Function Theorem

The Inverse Function Theorem states that if f is continuously differentiable and det f'(x) ≠ 0, then f is locally invertible near each point and the inverse function is continuously differentiable. Since f is 1-1, it is globally invertible.

Step 4: Substep 2.2 - Showing f(A) is Open

Since A is open and f is a homeomorphism (one-to-one, onto, and both f and f^{-1} are continuous), f(A) is open.

Step 5: Substep 2.3 - Differentiability of f^{-1}

By the Inverse Function Theorem, f^{-1} is continuously differentiable on f(A).

Step 3 - Proving Additional Property of f

To show that f(B) is open for any open set B ⊆ A, use the continuity of f and consider the image of open sets under continuous functions.

Step 7: Substep 3.1 - Continuity and Open Sets

If B is an open set in A, then since f is continuous, the image f(B) is an open set in f(A). Since f(A) is open, f(B) is open in ℝ^n.

Step 4 - Exercise 2-37(a)

Given: A continuously differentiable function f: ℝ^2 → ℝ. To show: f is not 1-1.

Step 9: Substep 4.1 - Using Rank Argument and Given Hint

If D₁f(x, y) ≠ 0 for all (x, y) in some open set A, defining g(x, y) = (f(x, y), y), notice g maps from ℝ^2 to ℝ^2 and its Jacobian matrix will have a non-zero determinant, meaning g can be 1-1 but f(x) = f(y) if x ≠ y in ℝ^2.

Step 5 - Exercise 2-37(b)

Generalize the above result to the case of a continuously differentiable function f: ℝ^n → ℝ^m where m

Step 11: Substep 5.1 - Rank Argument Generalization

In the more general scenario, construct a map g: ℝ^n → ℝ^n such that g(x) = (f(x), x_(m+1), ..., x_n). g's Jacobian matrix will have full rank (n), hence locally invertible, showing similar to above, f cannot be 1-1.

Step 6 - Exercise 2-38(a)

Given: f: ℝ → ℝ with f'(a) ≠ 0 for all a ∈ ℝ. To show: f is 1-1.

Step 13: Substep 6.1 - Monotonicity Implies Injectivity

If f'(a) > 0 for all a, f is strictly increasing and thus injective. Likewise, if f'(a) < 0 for all a, f is strictly decreasing, thus injective.

Step 7 - Exercise 2-38(b)

Given: f: ℝ^2 → ℝ^2 by f(x, y) = (e^x cos y, e^x sin y). To show: det f'(x, y) ≠ 0 for all (x, y) but f is not 1-1.

Step 15: Substep 7.1 - Computing the Jacobian

Calculate the Jacobian matrix:\[ \begin{pmatrix} \frac{∂}{∂x}(e^x \text{cos}(y)) & \frac{∂}{∂y}(e^x \text{cos}(y)) \ \frac{∂}{∂x}(e^x \text{sin}(y)) & \frac{∂}{∂y}(e^x \text{sin}(y)) \ \frac{∂}{∂x}(e^x \text{cos}(y)) = e^x \text{cos}(y), \frac{∂}{∂y}(e^x \text{cos}(y)) = -e^x \text{sin}(y), \frac{∂}{∂x}(e^x \text{sin}(y)) = e^x \text{sin}(y), \frac{∂}{∂y}(e^x \text{sin}(y)) = e^x \text{cos}(y) \ \text{Jacobian}_{f(x, y)} = e^x\begin{pmatrix} \text{cos}y & -\text{sin}y \ \text{sin}y & \text{cos}y \ \text{det(f)} = e^{2x}eq 0 \ \text{However, by analysing trigonometric behaviour}\ \text{(e.g.,} f(0, 0)=(1, 0), f(0, 2π) = (1, 0) and f is non- 1-1\text{as it can cycle back multiple inputs, similar to cos and sin being periodic.} \].

Key Concepts

continuously differentiable functions

When we talk about continuously differentiable functions, we mean functions that have continuous derivatives. This concept is crucial in calculus, especially when dealing with higher dimensions. A function \(f: \mathbf{R}^n \rightarrow \mathbf{R}^m\) is continuously differentiable if it has all partial derivatives \(\frac{\partial f}{\partial x_i}\) and they are continuous. This regularity ensures smooth changes of the function and forms the basis for many theorems, such as the Inverse Function Theorem. For example, in the exercise provided, \(f\) is continuously differentiable, guaranteeing that its derivative doesn't have abrupt changes.

open sets

Open sets are fundamental in topology and analysis, representing a set where, intuitively, you can move a little in any direction and still be in the set. More formally, a set \(A \subset \mathbf{R}^n\) is open if for every point \(x\in A\), there exists an \(\epsilon > 0\) such that the ball \(B(x, \epsilon) \subset A\). In the context of the exercise, proving that \(f(A)\) is open and \(f(B)\) is open for any open \( B\subset A\) relies on properties of open sets under continuous maps. Because \(f\) is continuously differentiable and has a continuously differentiable inverse, it ensures the behavior of open sets is preserved.

Inverse Function Theorem

The Inverse Function Theorem is a powerful tool in calculus. It states that if \(f: \mathbf{R}^n \rightarrow \mathbf{R}^n \) is continuously differentiable and \(\det(f'(x)) eq 0\) at some point \(x\), then \(f\) is locally invertible around \(x\). The inverse function \(f^{-1}\) will also be continuously differentiable around \(f(x)\). This theorem helps us in proving that \(f(A)\) is an open set and \(f^{-1}: f(A) \rightarrow A \) is differentiable in the provided exercise.

differentiability

Differentiability is a cornerstone in calculus. A function \(f\) is differentiable at \(x\) if it can be locally approximated by a linear map, which is its best linear approximation at that point. Mathematically, \(f\) is differentiable at \(x\) if there exists a linear map \(L\) such that \(\lim_{h\to 0}\frac{|f(x+h) - f(x) - L(h)|}{|h|} = 0\). In higher dimensions, this translates to the existence of a Jacobian matrix \(f'(x)\). For the function \(f\) in the exercise, differentiability ensures that various properties, like the behavior of images of open sets and the nature of inverse functions, can be smoothly analyzed.

Jacobian matrix

The Jacobian matrix is a matrix of all first-order partial derivatives of a vector-valued function. If \(f: \mathbf{R}^n \rightarrow \mathbf{R}^m\), then the Jacobian matrix is an \(m \times n\) matrix defined as \( J_f(x) = \left( \frac{\partial f_i}{\partial x_j} \right) \). It represents the best linear approximation of the function at a point. Its determinant, \(\det(J_f(x))\), plays a crucial role in determining the invertibility of \(f\). In the given exercise, \(\det(f'(x)) eq 0\) ensures that \(f\) is locally invertible and continuously differentiable using the Inverse Function Theorem.

injectivity

Injectivity is a property of a function where each element of the function's domain is mapped to a unique element in its codomain. A function \(f: A \rightarrow B\) is injective if \(f(x) = f(y)\) implies \(x = y\). In the context of the exercise, proving that \(f\) is 1-1 (injective) ensures that it has an inverse function. Injectivity is critical for establishing bijectivity, which along with continuity, helps prove that images of open sets are open.

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Problem 37

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Problem 39

Other exercises in this chapter

Problem 36

2-36.* Let \(A \subset \mathbf{R}^{n}\) be an open set and \(f: A \rightarrow \mathbf{R}^{n}\) a continuously differentiable \(1-1\) function such that \(\opera

View solution

Problem 37

2-37. (a) Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be a continuously differentiable function. Show that \(f\) is not 1-1. Hint: If, for example, \(D_{1}

View solution

Problem 39

2-39. Use the function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by $$ f(x)= \begin{cases}\frac{x}{2}+x^{2} \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0\end{ca

View solution

Problem 35

2-35. If \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) is differentiable and \(f(0)=0\), prove that there exist \(g_{i}: \mathbf{R}^{n} \rightarrow \mathbf{R}\)

View solution

Key Concepts

continuously differentiable functions

When we talk about continuously differentiable functions, we mean functions that have continuous derivatives. This concept is crucial in calculus, especially when dealing with higher dimensions. A function \(f: \mathbf{R}^n \rightarrow \mathbf{R}^m\) is continuously differentiable if it has all partial derivatives \(\frac{\partial f}{\partial x_i}\) and they are continuous. This regularity ensures smooth changes of the function and forms the basis for many theorems, such as the Inverse Function Theorem. For example, in the exercise provided, \(f\) is continuously differentiable, guaranteeing that its derivative doesn't have abrupt changes.

open sets

Open sets are fundamental in topology and analysis, representing a set where, intuitively, you can move a little in any direction and still be in the set. More formally, a set \(A \subset \mathbf{R}^n\) is open if for every point \(x\in A\), there exists an \(\epsilon > 0\) such that the ball \(B(x, \epsilon) \subset A\). In the context of the exercise, proving that \(f(A)\) is open and \(f(B)\) is open for any open \( B\subset A\) relies on properties of open sets under continuous maps. Because \(f\) is continuously differentiable and has a continuously differentiable inverse, it ensures the behavior of open sets is preserved.

Inverse Function Theorem

The Inverse Function Theorem is a powerful tool in calculus. It states that if \(f: \mathbf{R}^n \rightarrow \mathbf{R}^n \) is continuously differentiable and \(\det(f'(x)) eq 0\) at some point \(x\), then \(f\) is locally invertible around \(x\). The inverse function \(f^{-1}\) will also be continuously differentiable around \(f(x)\). This theorem helps us in proving that \(f(A)\) is an open set and \(f^{-1}: f(A) \rightarrow A \) is differentiable in the provided exercise.

differentiability

Differentiability is a cornerstone in calculus. A function \(f\) is differentiable at \(x\) if it can be locally approximated by a linear map, which is its best linear approximation at that point. Mathematically, \(f\) is differentiable at \(x\) if there exists a linear map \(L\) such that \(\lim_{h\to 0}\frac{|f(x+h) - f(x) - L(h)|}{|h|} = 0\). In higher dimensions, this translates to the existence of a Jacobian matrix \(f'(x)\). For the function \(f\) in the exercise, differentiability ensures that various properties, like the behavior of images of open sets and the nature of inverse functions, can be smoothly analyzed.

Jacobian matrix

The Jacobian matrix is a matrix of all first-order partial derivatives of a vector-valued function. If \(f: \mathbf{R}^n \rightarrow \mathbf{R}^m\), then the Jacobian matrix is an \(m \times n\) matrix defined as \( J_f(x) = \left( \frac{\partial f_i}{\partial x_j} \right) \). It represents the best linear approximation of the function at a point. Its determinant, \(\det(J_f(x))\), plays a crucial role in determining the invertibility of \(f\). In the given exercise, \(\det(f'(x)) eq 0\) ensures that \(f\) is locally invertible and continuously differentiable using the Inverse Function Theorem.

injectivity

Injectivity is a property of a function where each element of the function's domain is mapped to a unique element in its codomain. A function \(f: A \rightarrow B\) is injective if \(f(x) = f(y)\) implies \(x = y\). In the context of the exercise, proving that \(f\) is 1-1 (injective) ensures that it has an inverse function. Injectivity is critical for establishing bijectivity, which along with continuity, helps prove that images of open sets are open.