Problem 36
Question
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 \(\operatorname{det} f^{\prime}(x) \neq 0\) 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 \(B \subset A\).
Step-by-Step Solution
Verified Answer
By the Inverse Function Theorem, \( f(A) \) and \( f(B) \subset f(A) \) are open, and \( f^{-1} \) is differentiable.
1Step 1 - Understand the given function
We are given a continuously differentiable function \( f: A \rightarrow \mathbf{R}^{n} \) that is one-to-one (injective) on an open set \(A \subset \mathbf{R}^{n} \) with \( \operatorname{det} f^{\text{\textquotesingle}} (x) eq 0 \) for all \(x\).
2Step 2 - Apply Inverse Function Theorem
The Inverse Function Theorem states if a function \(f: U \rightarrow \mathbf{R}^{n} \) is continuously differentiable and \( \operatorname{det} f^{\text{\textquotesingle}} (x) eq 0 \) for all \(x \in U\), then \(f\) is locally invertible. This means there exists an open neighborhood \(V_x\) around each \(x \in A\) such that \(f\) maps \(V_x\) diffeomorphically onto its image \( f(V_x)\).
3Step 3 - Show that \(f(A)\) is open
Since \(f\) is locally invertible at each point in \(A\) and maps an open set to an open neighborhood of its image, the union of these neighborhoods over all points in \(A\) forms \( f(A) \). Therefore, \( f(A) \) is an open set.
4Step 4 - Prove \(f^{-1}\) is differentiable
Since \(f\) is continuously differentiable and locally invertible, the Inverse Function Theorem also guarantees that the inverse function \( f^{-1} \) is differentiable on \( f(A) \).
5Step 5 - Show that \(f(B)\) is open for any open set \(B \subset A\)
Let \(B\) be any open set within \(A\). Since \(f\) is injective and continuously differentiable, the same local invertibility argument applies. Each small open neighborhood within \(B\) is mapped to an open set in \( f(A) \), making \( f(B) \) open as it will be the union of these neighborhoods.
Key Concepts
Differentiable MapsInjectivityOpen Sets
Differentiable Maps
Differentiable maps are essential in understanding the behavior of functions in calculus. A function is differentiable if it has a well-defined derivative at every point in its domain. This means that the function behaves nicely, without abrupt changes in direction. Differentiable maps allow us to calculate how changes in input directly affect changes in output.
For a function from an open set in \( \bf{R}^n \) to \( \bf{R}^n \), continuous differentiability implies that not only does the derivative exist, but it varies smoothly across the domain. This is critical for applying the Inverse Function Theorem.
The given function \( f: A \rightarrow \bf{R}^{n} \) in the exercise adheres to these properties, ensuring that the function's behavior and the behavior of its inverse can be controlled and predicted precisely.
For a function from an open set in \( \bf{R}^n \) to \( \bf{R}^n \), continuous differentiability implies that not only does the derivative exist, but it varies smoothly across the domain. This is critical for applying the Inverse Function Theorem.
The given function \( f: A \rightarrow \bf{R}^{n} \) in the exercise adheres to these properties, ensuring that the function's behavior and the behavior of its inverse can be controlled and predicted precisely.
Injectivity
Injectivity, also known as one-to-one mapping, is when a function maps distinct elements from its domain to distinct elements in its range. This means no two different input values produce the same output. Injectivity is crucial for the existence of an inverse function.
In the exercise, the function \( f: A \rightarrow \bf{R}^{n} \) is injective. Therefore, for every point in the range, there is a unique corresponding point in the domain. Additionally, when combined with a non-zero determinant of the Jacobian matrix \(f'\textquotesingle(x) \), it sets the stage for the function to have a differentiable inverse (as guaranteed by the Inverse Function Theorem).
In the exercise, the function \( f: A \rightarrow \bf{R}^{n} \) is injective. Therefore, for every point in the range, there is a unique corresponding point in the domain. Additionally, when combined with a non-zero determinant of the Jacobian matrix \(f'\textquotesingle(x) \), it sets the stage for the function to have a differentiable inverse (as guaranteed by the Inverse Function Theorem).
Open Sets
Open sets are a cornerstone concept in topology and analysis. A set is open if, for every point in the set, there exists a small neighborhood around that point which is entirely contained within the set. Open sets are essential because they describe spaces where calculus tools like differentiation are well-behaved and applicable.
In the context of the exercise, the function \( f: A \rightarrow \bf{R}^{n} \) maps an open set \(A \) to another open set \( f(A) \). This property is significant because it means that the 'space' we are working with remains 'unbroken' and 'continuous' under the mapping. Additionally, the exercise requires you to show that the image of any open subset \(B \) under \(f \) is also an open set. This demonstrates that the function \(f \) preserves the open nature of sets within its domain.
In the context of the exercise, the function \( f: A \rightarrow \bf{R}^{n} \) maps an open set \(A \) to another open set \( f(A) \). This property is significant because it means that the 'space' we are working with remains 'unbroken' and 'continuous' under the mapping. Additionally, the exercise requires you to show that the image of any open subset \(B \) under \(f \) is also an open set. This demonstrates that the function \(f \) preserves the open nature of sets within its domain.
Other exercises in this chapter
Problem 34
2-34. A function \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) is homogeneous of degree \(m\) if \(f(t x)=\) \(t^{m} f(x)\) for all \(x .\) If \(f\) is also diff
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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 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 38
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 de
View solution