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I'm not sure if I can discuss some less math-majored problems here(i'm not a math-majored students), I'm studying convolution in the Fourier transform, and I see a sentence that convolution can be understood as using one function to smooth and average another function. I don't understand why it's smoother (I can probably feel some of it through some cases and through some thoughts, but I'd like to get a more robust explanation), and as I think about it deeply, I come across another question: how to define smoothness?

I have some rough explanations of g*f, such as using a sharp centered normalized g(x), and then I can find that the highest point of f(x) will fall, and the lowest point will rise, so the slope becomes smaller, so it is more smoother. However, I always feel the explanations are not strong enough, and I hope to have a better explanation.

or may be you have some excellent materials about convolutions with smooth and average can recommend to me?

Aerterliusi
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    Maybe smoothing a digital image by replacing the intensity value at each pixel with a weighted average of intensity values at nearby pixels would help build some intuition. – littleO Feb 08 '23 at 15:20
  • Convolving $f$ with $g$ corresponds to pointwise multiplication of $\hat f$ with $\hat g$. If $\hat g$ has bounded support, then multiplying by $\hat g$ kills off the high frequencies in $f$. So convolving with $g$ is a “low pass filter” operation. – littleO Feb 08 '23 at 15:26
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    Well, if $g(x)=1/N$ when $x\in [-N/2,N/2],$ and $0$ otherwise, then $f*g$ really is the moving average of $f.$ – Thomas Andrews Feb 08 '23 at 15:28
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    The convolution of $f$ with $g$ can be thought of as a weighted sum of shifted versions of $g$. So the result is at least as smooth as $g$ is. – littleO Feb 08 '23 at 15:28
  • So, convoluting with a step function is like a linear combination of sums of translated moving averages. And we can estimate most nice functions with step functions. – Thomas Andrews Feb 08 '23 at 15:32

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