1. ## Convolution problem

I'd like to prove that the only possible solution to $\displaystyle f = f*f$ with $\displaystyle f \in L^1$ is f = 0. So, I have the following;

1. Take the fourier transform of both sides

$\displaystyle \hat{f} = \widehat{f*f} = \sqrt{2\pi}\hat{f}\hat{f}$

2. Cancelling we get

$\displaystyle \hat{f} = \frac{1}{\sqrt{2\pi}}$

3. Writing the convolution outright and cancelling appropriately, we obtain

$\displaystyle \int e^{-ix\xi} f(x) dx = 1$

In $\displaystyle L^1$, this must mean that f = 0, given that the indefinite integral is constant.

2. Originally Posted by FLT
I'd like to prove that the only possible solution to $\displaystyle f = f*f$ with $\displaystyle f \in L^1$ is f = 0. So, I have the following;

1. Take the fourier transform of both sides

$\displaystyle \hat{f} = \widehat{f*f} = \sqrt{2\pi}\hat{f}\hat{f}$

2. Cancelling we get

$\displaystyle \hat{f} = \frac{1}{\sqrt{2\pi}}$

3. Writing the convolution outright and cancelling appropriately, we obtain

$\displaystyle \int e^{-ix\xi} f(x) dx = 1$

In $\displaystyle L^1$, this must mean that f = 0, given that the indefinite integral is constant.

Steps 1 and 2 are fine, but 3 looks unconvincing to me. Having got as far as step 2, I would use the Riemann–Lebesgue lemma, which says that $\displaystyle \hat{f}(\xi)\to0$ as $\displaystyle |\xi|\to\infty$, contradicting the assertion that $\displaystyle \hat{f} = \frac{1}{\sqrt{2\pi}}$. The contradiction comes from the cancellation in step 2, which assumed that $\displaystyle \hat{f}\ne0$. (So maybe step 2 wasn't quite as fine as I first said. )
Steps 1 and 2 are fine, but 3 looks unconvincing to me. Having got as far as step 2, I would use the Riemann–Lebesgue lemma, which says that $\displaystyle \hat{f}(\xi)\to0$ as $\displaystyle |\xi|\to\infty$, contradicting the assertion that $\displaystyle \hat{f} = \frac{1}{\sqrt{2\pi}}$. The contradiction comes from the cancellation in step 2, which assumed that $\displaystyle \hat{f}\ne0$. (So maybe step 2 wasn't quite as fine as I first said. )
If $\displaystyle f \in L^1$ then $\displaystyle \lim_{\xi\to\pm\infty}\hat{f}(\xi) = 0$