# How does a myopic interpret Wiener's Tauberian?

I just read about this post on the intuition behind convolution. In Terence Tao's answer convolution is interpreted as the blur of image in near-sighted eyes. In Harald Hanche-Olsen's it is made clearer that such a blur is due to the combining effect of translated images.

I wonder whether this interpretation can be carried on to explain Wiener's Tauberian theorem. I mean the following version

Suppose $\phi\in\mathcal{L}^{\infty}$, and $K\in\mathcal{L}^{1}$ be such that \begin{equation} lim_{x\to\infty} (\phi*K)(x)=a\hat{K}(0),\end{equation}and \begin{equation} \hat{K}(s)\neq 0\end{equation} for all $s$.

Then \begin{equation} lim_{x\to\infty} (\phi*f)(x)=a\hat{f}(0)\end{equation} for all $f\in\mathcal{L}^1$.

So the first equation describe the image in a near-sighted eye, and the last one says something about blurred image in other myopic eyes. But I do not know how to understand the right hand sight of both equations, and how the nonvanishing property relates to eye-sight.

Thanks!

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## 1 Answer

Well, $\hat K(0) = \int_{-\infty}^\infty K(x)\,\mathrm dx =: \int K$, so for a start perhaps one should divide both sides by $\int K$. Then we have $$\lim_{x\to\infty}\left(\phi*\frac K{\int K}\right)(x) = a.$$ Absorbing the division by $\int K$ into $K$ itself, it is sufficient to only consider those $K$ which are already normalized, i.e. which integrate to $1$. (Intuitively, the total amount of light entering your eyes is conserved, not amplified or diminished e.g. by sunglasses.) Similarly, we may do the same thing for $f$.

Then the theorem states the following: Suppose $\phi\in\mathcal L^\infty$ is an infinitely large image, and your blurry vision is described by a kernel $K\in\mathcal L^1$ such that:

1. $\int_{-\infty}^\infty K(x)\,\mathrm dx = 1$, i.e. you are not wearing sunglasses, and

2. $\hat K(s) \ne 0$ for any $s$, i.e. you can always tell a sinusoid from a featureless constant field. (This would not be true if, for example, your $K$ was a box function of width $w$ and $\phi$ was a sinusoid of wavelength exactly $w$: you would see it as a constant.)

Then the theorem states that, even though your perception of the image $\phi$ at any finite $x$ is poor, your perception of the limit as $x\to\infty$ is as good as anyone else's: $$\lim_{x\to\infty}(\phi*K)(x) = \lim_{x\to\infty}(\phi*f)(x)$$ for any $f\in\mathcal L^1$ with $\int f=1$. This feels like an intuitive result: no matter how big your blurring kernel is, it can't be so big that it corrupts your view of "$\phi(\infty)$" with values of $\phi(x)$ at any finite $x$. The worst that could happen is that there are never-ending oscillations to infinity that blur out to constant under your myopia, so you think the limit exists but it doesn't, but this is prevented by condition (2) above.

(Of course, if you or someone else is wearing sunglasses, all you have to do is account for the fact that you or they see the image as dimmer than usual, and scale the expected limit by the corresponding factor.)

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Good one! Thanks! – Hui Yu Oct 30 '12 at 7:59
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