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Problem 510

# Let $\mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}, \mathrm{G}(\mathrm{k})=\Phi\\{\mathrm{g}(\mathrm{x})\\}$ and suppose $\mathrm{F}(\mathrm{k}) \mathrm{G}(\mathrm{k})=\Phi\\{\mathrm{h}(\mathrm{x})\\}$. Prove the convolution theorem for Fourier transforms: If $$g(x)$$ and $$F(k)$$ are absolutely integrable on $(-\infty, \infty)$$and if the Fourier inversion integral for$$\mathrm{f}(\mathrm{x})$ is valid for all $$\mathrm{x}$$ except possibly a countably infinite number of points, then $$\mathrm{h}(\mathrm{x})=(\mathrm{f} * \mathrm{~g})$$ where ( $$f * g$$ ) is the convolution of $$f$$ and $$g$$ defined by $$(\mathrm{f} * \mathrm{~g})=\\{1 / \sqrt{(} 2 \pi)\\}^{\infty} \int_{-\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) \mathrm{d} \xi$$

Expert verified
In order to prove the convolution theorem for Fourier transforms, we first recall the definitions of the Fourier transform and convolution. Then, we substitute the given Fourier transforms F(k) and G(k) into their product and combine the integrals. Upon rearranging and comparing the resulting equation to the convolution definition, we can identify the similarity between the two expressions and conclude that h(x) = (f * g), thus proving the convolution theorem for Fourier transforms.
See the step by step solution

## Step 1: Recall the definition of the Fourier transform: $$\mathrm{F}(\mathrm{k})=\Phi\\{\mathrm{f}(\mathrm{x})\\}= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}$$ #Step 2: Convolution Definition#

Recall the definition of convolution: $$(\mathrm{f} * \mathrm{g})(\mathrm{x}) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) \, \mathrm{d}\xi$$ #Step 3: Starting with Fourier Transform Product#

## Step 2: We know that F(k)G(k) = Φ{h(x)}. Let's write this out: $$\mathrm{F}(\mathrm{k})\mathrm{G}(\mathrm{k}) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}$$ #Step 4: Substituting Fourier Transforms#

Now substitute the given Fourier transforms F(k) and G(k) in the equation from Step 3: $$\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}\right)\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{g}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}\right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}$$ #Step 5: Combining Integrals#

## Step 3: Now let's combine the product of the two integrals on the left side: $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) e^{-i \mathrm{k} \mathrm{x}} e^{-i \mathrm{k} \mathrm{\xi}} \, \mathrm{d}\xi \,\mathrm{d}\mathrm{x} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}$$ #Step 6: Rearrange and Compare to Convolution Definition#

Rewrite the left side of the equation from Step 5 using convolution definition: $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{f}(\xi) \mathrm{g}(\mathrm{x}-\xi) e^{-i \mathrm{k} \mathrm{\xi}} \, \mathrm{d}\xi \right) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \mathrm{h}(\mathrm{x}) e^{-i \mathrm{k} \mathrm{x}} \, \mathrm{d}\mathrm{x}$$ Now, we can identify the similarity between the left side of this equation and the convolution term defined in Step 2. Thus, we can conclude that h(x) = (f * g), which proves the convolution theorem for Fourier transforms.

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