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I have an $n \times n$ image $A$, and an $m\times m$ image $B$, where $n>m$. As the smaller image looks like a lower-resolution version of the larger one, I'm interested in the relative loss, computed as follows:

    1. use a Gaussian filter to low-pass $A$
    1. downsample the filtered $A$ to the size of $B$
    1. take the difference with $B$.

Specifically, let $f_\sigma$ be the Gaussian filter with bandwidth $\sigma$, and $\Phi(\cdot)$ be the function that downsamples every $n//m$ pixels along row and column. I want to compute $$\ell_\sigma = \lVert\Phi(f_\sigma \ast A) - B\rVert_F^2.$$

Here $\lVert\cdot\rVert_F$ is the Frobenius norm. $\ast$ is 2D convolution. I use subscript $\sigma$ because obviously the loss value depends on what bandwidth $\sigma$ we adopt.

I calculated the $\ell_\sigma$ for a range of $\sigma$'s. It seems $\ell_\sigma$ is a convex function of $\sigma$. Is that true for every $A$ and $B$? Is it possible to prove it?

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  • $\begingroup$ Is $n//m$ supposed to be $n/m$? $\endgroup$
    – LSpice
    Commented Jan 25, 2022 at 22:31
  • $\begingroup$ yes, I use the pythonic notation to indicate the rounding operation as necessary. But to simplify, you can assume n/m is an integer. $\endgroup$ Commented Jan 25, 2022 at 22:44

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