Suggested languages for you:

Americas

Europe

Problem 507

Prove the following properties of Fourier transforms: a) The Fourier transform of $$\mathrm{f}(\mathrm{x})$$ exists if $$\mathrm{f}$$ is absolutely integrable over $$(-\infty,+\infty)$$, b) If $$\mathrm{f}(\mathrm{x})$$ is real valued then $$F(-k)=F^{*}(k)$$ where $$F^{*}(k)$$ is the complex conjugate of $$F(k)$$.

Expert verified
In summary, we proved the following properties of Fourier transforms: a) The Fourier transform of f(x) exists if f is absolutely integrable over $$(-\infty,+\infty)$$. We showed this by utilizing the definition of absolute integrability and the Fourier transform, and proved that $$|F(k)| < \infty$$. b) If f(x) is real-valued, then $$F(-k) = F^{*}(k)$$, where $$F^{*}(k)$$ is the complex conjugate of $$F(k)$$. We used the definitions of the Fourier transform and complex conjugate to show that the expressions for F(-k) and F^{*}(k) are equal when f(x) is real-valued.
See the step by step solution

Step 1: Define absolute integrability

A function f(x) is absolutely integrable over an interval if: $$\int_{-\infty}^{\infty} |f(x)| dx < \infty$$ Using this, we will prove that the Fourier transform of f(x) exists.

Step 2: Apply the definition of the Fourier transform

We will use the definition of the Fourier transform, which states that: $$F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx$$

Step 3: Apply absolute integrability to the Fourier transform

If f(x) is absolutely integrable, then the following condition holds: $$\int_{-\infty}^{\infty} |f(x)| dx < \infty$$ We can rewrite the Fourier transform as follows: $$F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx = \int_{-\infty}^{\infty} f(x)(\cos(-2\pi k x)+i\sin(-2\pi k x)) dx$$ Now, we can split the integral into real and imaginary parts: $$F(k) = \int_{-\infty}^{\infty} f(x) \cos(-2\pi k x) dx + i \int_{-\infty}^{\infty} f(x) \sin(-2\pi k x) dx$$ Since the absolute value of a complex number is less than or equal to the sum of the absolute values of its real and imaginary parts, we have: $$|F(k)| \leq \int_{-\infty}^{\infty} |f(x)| |\cos(-2\pi k x)| dx + \int_{-\infty}^{\infty} |f(x)| |\sin(-2\pi k x)| dx$$ Both integrals on the right-hand side are absolutely integrable, since $$|\sin(-2\pi k x)| \leq 1$$ and $$|\cos(-2\pi k x)| \leq 1$$. Therefore, we can conclude that: $$|F(k)| < \infty$$ Since the Fourier transform is finite, we have proven that the Fourier transform of f(x) exists if f is absolutely integrable over $$(-\infty,+\infty)$$. ##Part b: Real-valued functions and complex conjugate## Now we will prove that if f(x) is real-valued, then $$F(-k) = F^{*}(k)$$.

Step 1: Rewrite the Fourier transform for F(-k)

We need to find the Fourier transform for F(-k), which is given by: $$F(-k) = \int_{-\infty}^{\infty} f(x) e^{2\pi i k x} dx$$

Step 2: Use the conjugate of the Fourier transform

From the definition of the complex conjugate: $$F^{*}(k) = \left(\int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} dx\right)^{*}$$ Now, we take the complex conjugate of the integral: $$F^{*}(k) = \int_{-\infty}^{\infty} f(x)^{*} \left(e^{-2\pi i k x}\right)^{*} dx$$ Given that f(x) is real-valued, we have $$f(x)^{*}=f(x)$$. Additionally, $$(e^{-2\pi i k x})^{*} = e^{2\pi i k x}$$, leading to: $$F^{*}(k) = \int_{-\infty}^{\infty} f(x) e^{2\pi i k x} dx$$

Step 3: Compare F(-k) and F^{*}(k)

Comparing the expressions for F(-k) and F^{*}(k), we found that for a real-valued function f(x): $$F(-k) = F^{*}(k)$$ This completes the proof of the property that if f(x) is real-valued, then the complex conjugate of $$F(k)$$ at -k is equal to $$F(-k)$$.

We value your feedback to improve our textbook solutions.

Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features

Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner