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Chapter 5: Problem 9

Cauchy-Schwarzsche Ungleichung. Es seien \(f, g: X \rightarrow \mathbb{K}\) me脽bar und \(|f|^{2},|g|^{2} \in\) \(\mathcal{L}^{1} .\) Zeigen Sie mit Hilfe des Satzes von FUBINI durch Betrachtung der Funktion \((x, y) \mapsto\) \(|f(x) g(x) f(y) g(y)|\) die Cauchy-Schwarzsche Ungleichung: $$ \left(\int_{X}|f g| d \mu\right)^{2} \leq\left(\int_{X}|f|^{2} d \mu\right)\left(\int_{X}|g|^{2} d \mu\right). $$ (Hinweis: Ist \(\mu\) nicht \(\sigma\)-endlich, so verschwinden \(f\) und \(g\) au脽erhalb einer me脽baren Menge \(\sigma\)-endlichen Ma脽es.)

Short Answer

Expert verified
The Cauchy-Schwarz inequality for integrable functions states that \((\int_{X}|fg| d\mu)^{2} \leq \int_{X}|f|^{2} d\mu \int_{X}|g|^{2} d\mu\), which is proven using Fubini's theorem.

Step by step solution

01

Establish useful relationships

From the given information it is known that \(|f|^{2},|g|^{2}\) are measurable and square integrable. Thus, the functions \(|f|\) and \(|g|\) are essentially bounded. We have to find a way to represent the integral of \(f(x)g(x)f(y)g(y)\) with the variables decoupled, which is necessary for applying the Fubini theorem.
02

Apply Fubini's Theorem

Now we rearrange the double integral. We can write \(\int_{X}\int_{X}|f(x)||g(x)||f(y)||g(y)|\,d\mu(x)\,d\mu(y) = \int_{X}\int_{X}|f(x)f(y)||g(x)g(y)|\,d\mu(x)\,d\mu(y)\). We can use Fubini's theorem to bring out common terms and rearrange the integral as \( \int_{X}\int_{X}|f(x)f(y)||g(x)g(y)|\,d\mu(x)\,d\mu(y) = \int_{X}|f(x)|^{2} d\mu(x) \int_{X}|g(y)|^{2} d\mu(y)\).
03

Prove the Cauchy-Schwarz Inequality

We now use the Cauchy-Schwarz inequality on the integrables \(|f(x)f(y)|\) and \(|g(x)g(y)|\), knowing that their integrals over X are finite. This yields the inequality \((\int_{X}|f(x)f(y)| d\mu(x) \int_{X}|g(x)g(y)| d\mu(y))^{2} \leq \int_{X}|f(x)|^{2} d\mu(x) \int_{X}|g(y)|^{2} d\mu(y)\). On replacing \(f(x)f(y)\) with \(fg\) (since the integral operation is linear) on the LHS, we get \((\int_{X}|fg| d\mu)^{2} \leq \int_{X}|f|^{2} d\mu \int_{X}|g|^{2} d\mu\), which completes the proof.

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Key Concepts

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

Measure Theory
Measure theory is a fascinating branch of mathematics focused on systematically defining and analyzing mathematical concepts like length, area, and volume. In our problem, it's crucial because it provides the foundation for understanding the integrals involved. Here are some key aspects:
  • Measurable Functions: These functions ensure that the values we obtain from given calculations are meaningful when integrating. In this context, \(|f|^{2}\) and \(|g|^{2}\) are measurable functions indicating they behave well under integration.
  • Measures: These are rules that assign a non-negative value to a set, which essentially quantifies its size in a generalized sense. A typical measure used is the Lebesgue measure, common in analyzing real-valued functions.
By confirming the functions involved are measurable, we ensure that further calculations, particularly those relating to integration, are sound and reliable.
Fubini's Theorem
Fubini's Theorem is a cornerstone in integration theory. It allows us to evaluate double integrals by integrating one variable at a time, simplifying complex calculations. This is particularly helpful in multidimensional integrals:
  • Double Integrals: In the context of our problem, we deal with a double integral constructed from the function \((x, y) \mapsto |f(x)g(x)f(y)g(y)|\). This can be perplexing without Fubini鈥檚 Theorem.
  • Switching Order: Fubini's Theorem tells us that under certain conditions, the order of integration can be exchanged, allowing computations to be carried out more conveniently. It shows that multi-layered integrations are often easier when done step-by-step.
By applying Fubini's Theorem in our proof, we rearrange and simplify the integral, a vital step towards validating the Cauchy-Schwarz inequality for integrals.
Integration Theory
Integration theory covers the methods of calculating the area under curves, along with more abstract constructs. It's pivotal to understanding how to solve integrals involving functions like \(f\) and \(g\) in our scenario:
  • Integrability: For our inequality, it's essential that the functions \(|f|^{2}\) and \(|g|^{2}\) are integrable over the space \(X\). This means that when we integrate them, we obtain a finite result which is crucial for further analysis.
  • Linear Properties: Integral operations are linear, allowing us to break down complex expressions into more manageable parts, aiding in the stepwise solution to our problem.
Through these principles of integration theory, we derive and solve integrals that are integral to proving inequalities like Cauchy-Schwarz, showing its wide applications.
Square Integrable Functions
Square Integrable Functions are functions whose square has a convergent integral. This is also known as \(L^2\) spaces in functional analysis. Understanding these functions is essential in our problem鈥檚 context:
  • \(L^1\) Spaces: While square integrable pertains to \(L^2\), the initial step is verifying they belong to \(L^1\) spaces (meaning they are absolutely integrable).
  • \(f, g\) as \(L^2\) Functions: The requirement is that both \(f\) and \(g\) are square integrable. This ensures that the product \(f \cdot g\) is meaningful and finite over the measure space, allowing the equality \(\int_{X}|fg| \,d\mu\) to be calculated.
Understanding square integrable functions is essential in signal processing, quantum mechanics, and various fields, underscoring their practical significance.

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Most popular questions from this chapter

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).)

Es sei \(\left(a_{n}\right)_{n \geq 1}\) eine Folge reeller Zahlen, und es gebe ein \(A \in \mathfrak{L}^{1}\) mit \(\lambda^{1}(A)>0\), so da脽 \(\lim _{n \rightarrow \infty} \exp \left(\overline{i a}_{n} x\right)\) f眉r alle \(x \in A\) existiert. Dann konvergiert die Folge \(\left(a_{n}\right)_{n \geq 1}\) in \(\mathbb{R}\). (Hinweise: Die Menge \(M\) der \(x \in \mathbb{R}\), f眉r welche \(g(x):=\lim _{n \rightarrow \infty} \exp \left(i a_{n} x\right)\) existiert, ist eine additive Gruppe. Nach dem Satz von STEINHAUS ist \(M=\mathbb{R}\). Eine Betrachtung von $$ \int_{\mathbb{R}} f(x) g(x) d x=\lim _{n \rightarrow \infty} \int_{\mathbb{R}} f(x) \exp \left(i a_{n} x\right) d x \quad\left(f \in \mathcal{L}^{1}(\mathbb{R})\right) $$ lehrt, \(\operatorname{da} 脽\left(a_{n}\right)_{n \geq 1}\) beschr盲nkt ist. Warum hat \(\left(a_{n}\right)_{n \geq 1}\) keine zwei verschiedenen H盲ufungspunkte?)

a) Es seien \(\alpha_{1}, \ldots, \alpha_{p}>0, Y:=\left\\{y \in \mathbb{R}^{p}: y>0, y_{1}+\ldots+y_{p}<1\right\\}\) und \(\left.f:\right] 0,1[\rightarrow[0, \infty]\) Borel-me脽bar. Dann gilt: $$ \int_{Y} f\left(y_{1}+\ldots+y_{p}\right) y_{1}^{\alpha_{1}-1} \cdot \ldots \cdot y_{p}^{\alpha_{p}-1} d \beta^{p}(y)=\frac{\Gamma\left(\alpha_{1}\right) \cdot \ldots \cdot \Gamma\left(\alpha_{p}\right)}{\Gamma\left(\alpha_{1}+\ldots+\alpha_{p}\right)} \int_{0}^{1} f(u) u^{\alpha_{1}+\ldots+\alpha_{p}-1} d u $$ und diese Gleichung gilt auch, falls \(f:] 0,1[\rightarrow \mathbb{K}\) Borel- me脽bar ist und eines der beiden Integrale existiert. (Hinweis: Benutzen Sie zur iterativen Berechnung des Integrals die Transformation \(t: X \rightarrow Y, t(x):=\left(x_{1}, \ldots, x_{p-2}, x_{p-1} x_{p}, x_{p-1}\left(1-x_{p}\right)\right)^{t}\), wobei \(X=\left\\{x \in \mathbb{R}^{p}: x>\right.\) \(\left.\left.0, x_{1}+\ldots+x_{p-1}<1, x_{p}<1\right\\} .\right)\) Ist zus盲tzlich \(\alpha_{p+1}>0\), so gilt: $$ \int_{Y}\left(1-\left(y_{1}+\ldots+y_{p}\right)\right)^{\alpha_{p+1}-1} y_{1}^{\alpha_{1}-1} \cdot \ldots \cdot y_{p}^{\alpha_{p}-1} d \beta^{p}(y)=\frac{\Gamma\left(\alpha_{1}\right) \cdot \ldots \cdot \Gamma\left(\alpha_{p+1}\right)}{\Gamma\left(\alpha_{1}+\ldots+\alpha_{p+1}\right)} $$ (DIRICHLET [1], S. 383 ff., [2], S. 375 ff.). b) Sind \(a_{1}, \ldots, a_{p}, \alpha_{1}, \ldots, \alpha_{p}, \beta_{1}, \ldots, \beta_{p}>0\) und \(Z:=\left\\{z \in \mathbb{R}^{p}: z>0,\left(z_{1} / a_{1}\right)^{\alpha_{1}}+\ldots+\right.\) \(\left.\left(z_{p} / a_{p}\right)^{\alpha_{p}}<1\right\\}, \rho_{j}:=\beta_{j} / \alpha_{j}(j=1, \ldots, p)\), so gilt unter entsprechenden Voraussetzungen an \(f:\) $$ \begin{aligned} &\int_{Z} f\left(\left(z_{1} / a_{1}\right)^{\alpha_{1}}+\ldots+\left(z_{p} / a_{p}\right)^{\alpha_{p}}\right) z_{1}^{\beta_{1}-1} \cdot \ldots \cdot z_{p}^{\beta_{p}-1} d \beta^{p}(z) \\ &=\frac{a_{1}^{\beta_{1}} \cdot \ldots \cdot a_{p}^{\beta_{p}}}{\alpha_{1} \cdot \ldots \cdot \alpha_{p}} \frac{\Gamma\left(\rho_{1}\right) \cdot \ldots \cdot \Gamma\left(\rho_{p}\right)}{\Gamma\left(\rho_{1}+\ldots+\rho_{p}\right)} \int_{0}^{1} f(u) u^{\rho_{1}+\cdots+\rho_{p}-1} d u \end{aligned} $$ c) Das Volumen des \(p\)-dimensionalen Ellipsoids \(E\left(a_{1}, \ldots, a_{p}\right):=\left\\{x \in \mathbb{R}^{p}:\left(x_{1} / a_{1}\right)^{2}+\ldots+\right.\) \(\left.\left(x_{p} / a_{p}\right)^{2}<1\right\\}\) betr盲gt $$ \beta^{p}\left(E\left(a_{1}, \ldots, a_{p}\right)\right)=\frac{\pi^{p / 2}}{\Gamma\left(\frac{p}{2}+1\right)} a_{1} \cdot \ldots \cdot a_{p} $$ speziell ist $$ \beta^{p}\left(K_{r}(0)\right)=\frac{\pi^{p / 2}}{\Gamma\left(\frac{p}{2}+1\right)} r^{p}. $$

Es sei \(f_{\alpha}(x, y):=x \cdot y /\left(x^{2}+y^{2}+1\right)^{\alpha} \quad(x, y \in \mathbb{R})\). Bestimmen Sie alle \(\alpha \in \mathbb{R}\), f眉r welche die iterierten Integrale \(\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty} f_{\alpha}(x, y) d x\right) d y\) und \(\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty} f_{\alpha}(x, y) d y\right) d x\) existieren. F眉r welche \(\alpha\) ist \(f_{\alpha} \beta^{2}\)-integrierbar 眉ber \(\mathbb{R}^{2} ?\)

Mit \(M:=\left\\{(x, y)^{t} \in \mathbb{R}^{2}: x0\right\\}\) gilt: $$ \int_{M} y e^{-\frac{1}{2}\left(x^{2}+y^{2}\right)} d \beta^{2}(x, y)=\frac{1}{2}(1+\sqrt{2}) \sqrt{\pi}. $$
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