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Suppose that \(U\) and \(V\) are open in \(R^{n}\) and \(\Phi: V \rightarrow U\) is a \(C^{\infty}\) diffeomorphism. Explain how to define \(F \circ \Phi \in \mathcal{D}^{\prime}(U)\) for any \(F \in \mathcal{D}^{\prime}(V)\).

Short Answer

Expert verified
The pullback distribution \( F \circ \Phi \) acts as \((F \circ \Phi)(\varphi) = F(\varphi \circ \Phi)\) for \( \varphi \in C_c^\infty(U) \).

Step by step solution

01

Understand the Function Mapping

We have a smooth map \( \Phi: V \rightarrow U \) which is a \( C^\infty \) diffeomorphism. This means that \( \Phi \) is smooth, bijective, and has a smooth inverse. Our goal will be to show how a distribution \( F \) on \( V \) can be transformed into a distribution on \( U \) via this mapping.
02

Define Distributions on Manifolds

A distribution \( F \in \mathcal{D'}(V) \) acts on smooth compactly-supported functions on \( V \), denoted \( C_c^\infty(V) \). To find \( F \circ \Phi \), we will need to consider how this distribution interacts with test functions on \( U \).
03

Develop the Pullback Mechanism

Given a test function \( \varphi \in C_c^\infty(U) \), we can pull it back to \( V \) using \( \Phi \). Create a new function \( \varphi \circ \Phi \) which will be defined on \( V \). Since \( \Phi \) is a diffeomorphism, \( \varphi \circ \Phi \) is smooth and has compact support.
04

Define the Action of the Pullback Distribution

Using the pullback mechanism, define the action of \( F \circ \Phi \) on a test function \( \varphi \) from \( U \) as follows: \((F \circ \Phi)(\varphi) = F(\varphi \circ \Phi)\). This defines a distribution on \( U \) by mapping \( \varphi \) to the action of \( F \) on the pullback \( \varphi \circ \Phi \).
05

Verify the Definition

Check that this definition respects the linearity and continuity of distributions. Since \( \Phi \) and its inverse are smooth and \( \varphi \) is smooth compactly-supported, it means that \( \varphi \circ \Phi \) remains in \( C_c^\infty(V) \), preserving the required properties for \( F(\varphi \circ \Phi) \) to be a valid action of a distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Smooth map
In the context of mathematics, a 'smooth map' is a type of function that is infinitely differentiable. When we say that a function is smooth, we often denote it as a function of class \( C^{\infty} \). This means the function can be differentiated any number of times, and each derivative is a continuous function of its own.

A key feature of smooth functions is that they have no sharp edges or discontinuities. Imagine a beautiful curve flowing gently without any jagged turns—this is akin to what a smooth function looks like in mathematical terms.
  • **Bijective Smooth Map:** In our original problem, \( \Phi: V \rightarrow U \) is described as a \( C^{\infty} \) diffeomorphism. This indicates that \( \Phi \) is not only smooth but also bijective (one-to-one and onto), implying \( \Phi \) has a smooth inverse.
  • **Role in Diffeomorphism:** The smoothness ensures that we can pull back or push forward other functions along with \( \Phi \) and its inverse without losing their smooth nature. This comes into play particularly when dealing with distributions (explained below).
Distribution
In mathematical analysis, particularly on a manifold like \( V \), a 'distribution' is a generalization of a function. These provide a way to handle functions that might be too jumpy or not well-behaved for classical calculus. Distributions, denoted as \( \mathcal{D}^\prime \), extend the idea of functions by allowing us to work with objects that act on test functions.

Distributions do not represent pointwise functions but work by assigning numbers to test functions. A key aspect of distributions is that they act on smooth, compactly-supported test functions.
  • **Distributions on \( V \):** In the problem, a distribution \( F \in \mathcal{D}^\prime (V) \) acts on test functions from \( V \), allowing a broader view since these test functions are smooth with compact support.
  • **Transformation via \( \Phi \):** With the map \( \Phi \), and smooth test functions, we need to establish how we can transform \( F \) from \( V \) to \( U \) by utilizing \( \Phi \). This transformation is crucial as it ensures the test functions remain within the space \( C_c^\infty(V) \).
Thus, distributions offer robust tools to handle mathematical entities that are beyond classical function interpretation.
Compact support
The term 'compact support' is a property of functions in mathematics. Functions that are compactly supported are zero outside a certain bounded region and are crucial when dealing with distributions and integrals.

These functions help in ensuring that we work with a finite space, which simplifies many complex mathematical operations. This bounded nature is what breaks infinitesimally vast spaces into manageable segments that do not blow up to infinity or linger indefinitely.
  • **Smooth Compactly-Supported Functions:** For our task of pulling back, if \( \varphi \) is in \( C_c^\infty(U) \), its pullback \( \varphi \circ \Phi \) is guaranteed to have compact support, making it smoother and more workable.
  • **Importance in Transformations:** The concept of compact support ensures that the distribution transformations respect necessary conditions, keeping all calculations within contained, bounded regions.
Compact support not only preserves mathematical rigor but simplifies many explanations, providing clarity in solving complex real-world problems.

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Most popular questions from this chapter

Suppose that \(f_{1}, f_{2}, \ldots\), and \(f\) are in \(L_{\text {loc }}^{1}(U) .\) The conditions in (a) and (b) below imply that \(f_{n} \rightarrow f\) in \(\mathcal{D}^{\prime}(U)\), but the condition in (c) does not. a. \(f_{n} \in L^{P}(U)(1 \leq p \leq \infty)\) and \(f_{n} \rightarrow f\) in the \(L^{P}\) norm or weakly in \(L^{P}\). b. For all \(n,\left|f_{n}\right| \leq g\) for some \(g \in L_{\text {loe }}^{1}(U)\), and \(f_{n} \rightarrow f\) a.e. c. \(f_{n} \rightarrow f\) pointwise.

Let \(F\) be a distribution on \(\mathbb{R}^{n}\) such that \(\operatorname{supp}(F)=\\{0\\}\). a. There exist \(N \in \mathbb{N}, C>0\) such that for all \(\phi \in C_{e}^{\infty}\). $$ |\langle F, \phi\rangle| \leq C \sum_{|a| \leq N} \sup _{|x| \leq 1}\left|\partial^{\alpha} \phi(x)\right| $$ b. Fix \(\psi \in C_{e}^{\infty}\) with \(\psi(x)=1\) for \(|x| \leq 1\) and \(\psi(x)=0\) for \(|x| \geq 2\). If \(\phi \in C_{e}^{\infty}\), let \(\phi_{k}(x)=\phi(x)[1-\psi(k x)]\). If \(\partial^{\alpha} \phi(0)=0\) for \(|\alpha| \leq N\), then \(\partial^{\alpha} \phi_{k} \rightarrow \partial^{\alpha} \phi\) uniformly as \(k \rightarrow \infty\) for \(|\alpha| \leq N\). (Hint: By 'Taylor's theorem. \(\left|\partial^{\alpha} \phi(x)\right| \leq C|x|^{N+1-|\alpha|}\) for \(\left.|\alpha| \leq N_{.}\right)\) c. If \(\phi \in C_{c}^{\infty}\) and \(\partial^{\alpha} \phi(0)=0\) for \(|\alpha| \leq N\), then \(\langle F, \phi\rangle=0\). d. There exist constants \(c_{\alpha}(|\alpha| \leq N)\) such that \(F=\sum_{|a| \leq N} c_{\alpha} \partial^{\alpha} \delta\).

Let \(f\) be a continuous function on \(\mathbb{R}^{n} \backslash\\{0\\}\) that is homogeneous of degree \(-n\) (i.e., \(\left.f(r x)=r^{-n} f(x)\right)\) and has mean zero on the unit sphere (i.e., \(\int f d \sigma=0\) where \(\sigma\) is surface measure on the sphere). Then \(f\) is not locally integrable near the origin (unless \(f=0)\), but the formula $$ \langle P V(f), \phi\rangle=\lim _{\epsilon \rightarrow 0} \int_{|x|>\epsilon} f(x) \phi(x) d x \quad\left(\phi \in C_{c}^{\infty}\right) $$ defines a distribution \(P V(f)-" P V "\) stands for "principal value" - that agrees with \(f\) on \(\mathbb{R}^{n} \backslash\\{0\\}\) and is homogeneous of degree \(-n\) in the sense of Exercise 9 .

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