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I have to show that the convolution of a function $f \in L^1(\mathbf{R})$ with the harmonic oscillation $\phi_\omega (t) = \exp(2 \pi i t \omega)$ is equal to the Fourier Transform of $f$, $\hat{f}(\omega)$:

$$ f * \phi_\omega = \hat{f} (\omega) = \int_{-\infty}^{\infty} f(t) \exp(-2 \pi i t \omega) dt $$

However I cannot quite get it...

Attempt 1:
I tried to show it using the definition of the convolution

$$ f * g = \int_{-\infty}^{\infty} f(\tau)g(t-\tau) d \tau, $$

this gives me (by exchanging the role of $t$ and $\tau$ in the above equation)

$$ \int_{-\infty}^{\infty} f(t)\exp(2 \pi i (\tau - t) \omega) dt = \exp(2 \pi i \tau \omega) \int_{-\infty}^{\infty} f(t)\exp(-2 \pi i t \omega) dt = \exp(2 \pi i \tau \omega) \hat{f}(\omega) $$

but this is the fourier transform of $f(t+\tau)$, right?

Attempt 2: Using the convolution theorem

$$ \mathcal{F}(f * g) = \mathcal{F}(f) \cdot \mathcal{F}(g) $$

we have

$$ f * g = \mathcal{F}^{-1}\mathcal{F}(f * \phi_\omega) = \mathcal{F}^{-1}(\mathcal{F}(f) \cdot \mathcal{F}(\phi_\omega)) $$

where $\mathcal{F}(f) = \hat{f}(\omega)$ and $\mathcal{F}(\phi_w) = \hat{\phi_w}(\omega) = \delta(\omega-w)$?

Then we have

$$ f * g = \mathcal{F}^{-1}(\hat{f}(\omega) \delta(\omega-w)) = \int_{-\infty}^{\infty}\hat{f}(\omega) \delta(\omega-w) \exp(2 \pi i t \omega) d\omega $$

And since there only is a nonzero contribution for $\omega = w$ this gives

$$ f * g = \exp(2 \pi i t w) \hat{f}(\omega) $$

I do have the feeling I messed up some variables and constants :(
What am I missing, or doing wrong?

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The Fourier transform of f is a function of $\omega$ whereas the convolution is a function of both $\omega$ and $t$... – yoki Dec 26 '12 at 11:53
Perhaps the problem was meant to be about "convolution of $f$ with $\phi_\omega$, evaluated at $t=0$"? – user53153 Dec 27 '12 at 4:29
@ido could you elaborate please? I already was confused by the $\omega$ in $\phi_\omega(t)$ and how to treat it in attempt 2. Are you saying the problem doesn't make sense? – Gerhard Burger Dec 27 '12 at 8:24
In my understanding, you wish to prove that the fourier transform, which is a function of $\omega$ alone, is equal to the convolution (in the time domain) of two functions, where one has a parameter $\omega$. So you wish to prove that a function of $\omega$ is equal to a function of $\omega$ and $t$. Since $\omega$ is clearly affecting the result, I don't see how this can be correct. – yoki Dec 27 '12 at 10:37
@ido Thanks, I will ask my professor for clarification – Gerhard Burger Dec 27 '12 at 10:47

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