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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 $$F(k)$$.

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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 $$F(-k)$$.

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