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Chapter 5: Problem 4

In der Situation des Satzes von PLANCHEREL gilt f眉r alle \(x \in \mathbb{R}^{p}\) : $$ \int_{\mathbb{R}^{p}} f(t+x) \overline{f(t)} d \mu_{p}(t)=\int_{\mathbb{R}^{p}}|\hat{f}(t)|^{2} e^{i\langle t, x\rangle} d \mu_{p}(t) $$ (Hinweis: Grenz眉bergang \(n \rightarrow \infty\) in (3.8) oder Parsevalsche Formel.)

Short Answer

Expert verified
By substituting \( f(t) \) with \( f(t+x) \) in Parseval's formula, and carrying out the integral according to the properties of Fourier transforms, we are able to show the correctness of the given equation as per the theorem of Plancherel.

Step by step solution

01

Understanding the given equation

The given equation is of the form \[\int_{\mathbb{R}^{p}} f(t+x) \overline{f(t)} d \mu_{p}(t)=\int_{\mathbb{R}^{p}}|\hat{f}(t)|^{2} e^{i\langle t, x\rangle} d \mu_{p}(t) \]where \( f \) is a function and \( \hat{f} \) is its Fourier Transform. \( \mu_{p} \) is the Lebesgue measure on \( \mathbb{R}^{p} \). \( \overline{f(t)} \) represents the complex conjugate of \( f(t) \).
02

Apply Parseval's Formula

Parseval's formula is a result that connects the Fourier transform with the original function. It is given as\[\int_{\mathbb{R}^{p}} |f(t)|^{2} d \mu_{p}(t) = \int_{\mathbb{R}^{p}} |\hat{f}(t)|^{2} d \mu_{p}(t)\]Applying Parseval's formula, we substitute \( f(t) \) with \( f(t+x) \) which gives \[\int_{\mathbb{R}^{p}} |f(t+x)|^{2} d \mu_{p}(t) = \int_{\mathbb{R}^{p}} |\hat{f}(t)|^{2} e^{i\langle t, x\rangle} d \mu_{p}(t)\]
03

Apply the property of the Fourier Transform

To prove that this is true, we remember a property of the Fourier transform. Using this property, we know that \[\hat{f}(t+x)=\hat{f}(t)e^{i\langle t, x\rangle}\]Hence \[\int_{\mathbb{R}^{p}} |f(t+x)|^{2} d \mu_{p}(t) = \int_{\mathbb{R}^{p}} |\hat{f}(t)|^{2} e^{i\langle t, x\rangle} d \mu_{p}(t)\]is true, which completes the proof.

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Key Concepts

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

Fourier Transform
The Fourier Transform is a powerful tool in mathematics and signal processing that transforms a function of time into a function of frequency. This means it takes a time-domain signal and turns it into a frequency-domain representation. For a function \( f(t) \), the Fourier Transform \( \hat{f}(t) \) is defined as:\[\hat{f}(\xi) = \int_{\mathbb{R}^{p}} f(t) e^{-i\langle t, \xi \rangle} d\mu_p(t)\]
  • It expresses the original function in terms of its frequency components.
  • Used in various fields such as physics, engineering, and finance.
  • Helps analyze and understand signals and their frequency behaviors.
By using the Fourier Transform, we can study the properties of signals and functions in a different light, helping us to gain insights into their underlying patterns and characteristics. When dealing with complex numbers, don't forget the influence of the imaginary unit \( i\) and the importance of both amplitude and phase in representing your signal.
Parseval's Formula
Parseval's Formula creates a bridge between a function and its Fourier Transform by showing that the total energy of the function in time space is equal to the total energy in frequency space. It is represented as follows:\[\int_{\mathbb{R}^{p}} |f(t)|^{2} d \mu_p(t) = \int_{\mathbb{R}^{p}} |\hat{f}(t)|^{2} d \mu_p(t)\]This tells us that the integral of the square of the function over all space (representing "energy" or "power") remains constant whether it's in the time domain or Fourier domain.
  • Useful in signal processing as it relates energy conservation between time and frequency domains.
  • Allows evaluation of functions in a more convenient domain.
  • Serves as the foundation to identify symmetries and invariants in physical systems.
Parseval's Formula is invaluable for understanding how different functions behave under transformation and ensuring their consistent properties across domains.
Lebesgue Measure
The Lebesgue Measure \( \mu_p \), is a mathematical concept that allows for defining and analyzing the size or measure of sets, particularly in high-dimensional spaces like \( \mathbb{R}^p \). Unlike the standard notion of length or area, it is more adaptable and can handle more general sets.
  • Enables integration over more complex sets and functions.
  • Crucial for formalizing the concept of an integral, known as the Lebesgue Integral.
  • Serves as a fundamental tool in modern real-analysis and probability theory.
The Lebesgue Measure is essential for accurate and meaningful integration within the framework of analysis on \( \mathbb{R}^p \), and it underlies the proper functioning of the Fourier Transform and Parseval's Formula by ensuring these operations are well-defined and extend to more general functions and domains.

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Most popular questions from this chapter

F眉r zwei \(\sigma\)-Ringe \(\Re\) 眉ber \(X, \mathfrak{S}\) 眉ber \(Y\) sei \(\Re \otimes \mathfrak{S}\) der von \(\Re *(\mathfrak{S}\) erzeugte \(\sigma\)-Ring 眉ber \(X \times Y . \mathfrak{N}_{\mu}\) sei der \(\sigma\)-Ring aller Teilmengen von \(\mu\)-Nullmengen. a) F眉r alle \(P \in \mathfrak{A} \otimes \mathfrak{S}, N \in \mathfrak{N}_{\mu} \otimes \mathfrak{S}\) gilt: \(P \cap N \in \mathfrak{N}_{\mu} \otimes \mathfrak{S}\). b) \(\tilde{\mathfrak{A}} \otimes \mathfrak{S}=\left\\{P \cup N: P \in \mathfrak{A} \otimes \mathfrak{S}, N \in \mathfrak{N}_{\mu} \otimes \mathfrak{S}\right\\}\) c) \(\tilde{\mathfrak{A}} \otimes \tilde{\mathfrak{B}}=\left\\{P \cup A \cup B \cup C: P \in \mathfrak{A} \otimes \mathfrak{B}, A \in \mathfrak{N}_{\mu} \otimes \mathfrak{B}, B \in \mathfrak{A} \otimes \mathfrak{N}_{\nu}, C \in \mathfrak{N}_{\mu} \otimes \mathfrak{N}_{\nu}\right\\}\).

a) F眉r \(A \in \operatorname{GL}(p, \mathbb{R})\) ist $$ \int_{\mathbb{R}^{p}} e^{-\|A x\|^{2}} d \beta^{p}(x)=\pi^{p / 2}|\operatorname{det} A|^{-1} $$ b) Ist \(A \in \mathrm{GL}(p, \mathbb{R})\) positiv definit, so gilt: $$ \int_{\mathbb{R}^{p}} e^{-\langle A x, x\rangle} d \beta^{p}(x)=\pi^{p / 2}(\operatorname{det} A)^{-1 / 2}. $$

F眉r \(M \subset \mathbb{R}^{p}\) sei card \(M \in[0, \infty]\) die Anzahl der Elemente von \(M .\) Es gilt f眉r alle \(M \in \mathfrak{B}^{p}\) : Die Funktion \(x \mapsto \operatorname{card}\left((M+x) \cap \mathbb{Z}^{p}\right)\) ist Borel- me脽bar und $$ \beta^{p}(M)=\int_{[0,1[p} \operatorname{card}\left((M+x) \cap \mathbb{Z}^{p}\right) d \beta^{p}(x) $$ Ist also \(\beta^{p}(M)>1(\) bzw. \(<1)\), so existiert eine Borel-Menge \(A \subset\left[0,1\left[^{p}\right.\right.\) mit \(\beta^{p}(A)>0\), so \(\operatorname{da} \mathfrak{c a r d}\left((M+x) \cap \mathbb{Z}^{p}\right) \geq 2(\) bzw. \(=0)\) f眉r alle \(x \in A\). Entsprechendes gilt f眉r \(\lambda^{p}\) statt \(\beta^{p}\) (Bemerkung: Von H. STEINHAUS stammt folgendes Problem: Gibt es eine Menge \(M \subset \mathbb{R}^{p}\), so da脽 \(\operatorname{card}\left(t(M) \cap \mathbb{Z}^{2}\right)=1\) f眉r jede Bewegung \(t: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} ?-\) Es ist bekannt, da脽 keine beschr盲nkte Menge \(M \in \mathfrak{L}^{2}\) das Gew眉nschte leistet; s. J. BECK: On a lattice-point problem of \(H .\) Steinhaus, Stud. Sci. Math. Hung. 24, 263-268 (1989); s. auch P. KomJ脕TH: A latticepoint problem of Steinhaus, Quart. J. Math., Oxf. (2) 43, 235-241 (1992).)

F眉r \(n \geq 1\) sei \(E_{n}:=\left\\{x \in \mathbb{R}^{n}:\|x\|<1\right\\} .-\) Es seien nun \(p \geq 2\) und \(X:=\) ] \(0, \infty\left[\times E_{p-1}, \bar{Y}:=\right] 0, \infty\left[\times \mathbb{R}^{p-1}, t: X \rightarrow Y\right.\) $$ t(r, x):=r\left(\left(1-\|x\|^{2}\right)^{1 / 2}, x\right) \quad\left(r>0, x \in E_{p-1}\right) $$ Dann ist \(t\) ein \(C^{1}\)-Diffeomorphismus mit det \(D t(r, x)=r^{p-1}\left(1-\|x\|^{2}\right)^{-1 / 2}\). Ist \(\left.F:\right] 0, \infty[\rightarrow \hat{K}\) Borel-me脽bar und \(F(r) r^{p-1}\) 眉ber \(] 0, \infty\left[\beta^{1}\right.\)-integrierbar, so gilt: $$ \int_{\mathbb{R}^{p}} F(\|y\|) d \beta^{p}(y)=2\left(\int_{0}^{\infty} F(r) r^{p-1} d r\right) \cdot \int_{E_{p-1}}\left(1-\|x\|^{2}\right)^{-1 / 2} d \beta^{p-1}(x) $$ Insbesondere resultiert f眉r \(F=\chi_{] 0,1} \mid\) $$ \int_{E_{p-1}}\left(1-\|x\|^{2}\right)^{-1 / 2} d \beta^{p-1}(x)=\frac{p}{2} \beta^{p}\left(E_{p}\right) $$ und f眉r \(F(r)=\exp \left(-r^{2}\right):\) $$ \beta^{p}\left(E_{p}\right)=\frac{\pi^{p / 2}}{\Gamma\left(\frac{p}{2}+1\right)} $$

F眉r abz盲hlbare Mengen \(X, Y\) gilt: \(\mathfrak{P}(X) \otimes \mathfrak{P}(Y)=\mathfrak{P}(X \times Y)\). Ist dagegen \(|X|>|\mathbb{R}|\), so ist \(\mathfrak{P}(X) \otimes \mathfrak{P}(X)\) eine echte Teilmenge von \(\mathfrak{P}(X \times X)\). (Hinweis: Korollar III.5.15. Bemerkung: Unter Annahme der Kontinuumshypothese ist \(\mathfrak{P}(X) \otimes \mathfrak{P}(X)=\mathfrak{P}(X \times X)\), falls \(|X| \leq|\mathbb{R}| ;\) s. B.V. RAO: On discrete Borel spaces and projective sets, Bull. Amer. Math. Soc. \(75,614-617\) (1969) und A.B. KHARAZISHVILI: A note on the Sierpi?ski partition, J. Appl. Anal. 2, 41-48 (1996).)
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