### Author Topic: HA7-P1  (Read 2498 times)

#### Victor Ivrii

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##### HA7-P1
« on: November 01, 2015, 05:06:30 PM »
« Last Edit: November 01, 2015, 05:09:09 PM by Victor Ivrii »

#### Emily Deibert

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##### Re: HA7-P1
« Reply #1 on: November 03, 2015, 01:47:02 PM »
(b) We start by considering $(F^2f)(x) = F(Ff)(x)$.

We have that:
F(f(x)) = \hat{f}(k) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} f(x)e^{-ikx}dx

Now consider that:
f(x) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} \hat{f}(k)e^{ikx}dk

Then
F^2(f(x)) = F(F(f(x)) = F(\hat{f}(k)) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} \hat{f}(k)e^{-ikx}dk = f(-x)

Since an even function is defined such that $f(-x) = f(x)$, the above operation will recover the original function. For an odd function, which has the property that $f(-x) = -f(x)$, the operation will recover the negative of the function.
« Last Edit: November 03, 2015, 01:49:35 PM by Emily Deibert »

#### Emily Deibert

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##### Re: HA7-P1
« Reply #2 on: November 03, 2015, 01:54:02 PM »
(c) We want to show that applying the Fourier transform operator four times will recover the original function (i.e. the fourth power of the Fourier transform operator is equivalent to the Identity). This is easily built off of part (b). For consider the result of part (b):

F^2(f(x)) = f(-x)

Now let's apply this same result again:

F^2[F^2(f(x))] = F^2(f(-x)) = f(-(-x)) = f(x) \longrightarrow F^4(f(x)) = f(x)

Since this operation recovers the original function for all functions, it must be that the fourth power of the Fourier transform operator is the Identity:

F^4 = I

#### Rong Wei

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##### Re: HA7-P1
« Reply #3 on: November 04, 2015, 04:57:56 PM »
for question dï¼‰

#### Emily Deibert

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##### Re: HA7-P1
« Reply #4 on: November 05, 2015, 10:37:35 AM »
Professor, would you be able to provide a hint for part (a)? I'm not sure how to get started on it.

#### Victor Ivrii

If $Fu = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-ikx}u(x)\,dx$ then we know that $\| Fu\|^2= \|u\|^2$ for all $u$ and therefore $F^*F=I$. For infinite-dimensional operators it alone does not imply that $FF^*=I$ (only that $FF^*$ is an orthogonal projector) but since we know that $F$ is invertible  $F^*F=I\implies F^*F F^{-1}= F^{-1}\implies F^*=F^{-1}\implies FF^*=FF^{-1}=I$.