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I have asked this question a while back on StackExchange but have not received any answer/comment. I received a suggestion to post the same question in here which is more research oriented.

Let $k*f(x)=\int_{}^{}k(x,t)f(x-t)dt$, where $x,t\in \mathbb{R}^{3}$, $f:\mathbb{R}^{3}\to\mathbb{R^{+}}$ and $k:\mathbb{R}^{3}\to\mathbb{R^{+}}$ for a given $t$. Additionally, let $v(x):\mathbb{R}^{3}\to\mathbb{R}^{3}$ and $D_{v(x)}(f(x))$ as directional derivative of $f(x)$ along $v(x)$. Is it possible to write $k*D_{v}(f)$ as a function of $k*f$:

$$k*D_{v}(f)=F(k*f)$$

where $k*D_{v}(f)=\int{}^{}k(x,t)v(x-t)\nabla^{T}f(x-t)dt.$ Or is there any map/operator $F$ that generates $k*D_{v}(f)$ by acting on $k*f$ ?

In other words, the objective is finding $F$ in a way that path (1) and (2), in the following diagram, lead to the same function. enter image description here

Note1: In my problem, $f$ is unknown but $k$, $v$ and $k*f$ are known numerically. Therefore, in addition to $k*f$, the map $F$ can be dependent to $k$ and/or $v$ and/or $k*v$ and/or their derivatives but there should not be any dependency to $f$ or its derivatives.

Note2: I must say that I am not a mathematician and would like to apologize beforehand if my mathemtical language here and in the comment section is impercise. I welcome any comments for improvement. Thank you.

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    $\begingroup$ I'd say "equivalent" rather than "equivariant." The StackExchange link goes to your figure, not a question on another site. $\endgroup$
    – Brian Hopkins
    Commented Mar 3, 2022 at 17:42

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You will need the inverse kernel $q(x,t)$ of $k(x,t)$, such that $$\int q(y,t)k(x,t)\,dt=\delta(x-y).$$ This can be obtained by Fourier transformation of the known function $k(x,t)$. Then the desired function $F$ is given by $$F\cdots=k\ast \bigl(Dv \,(q\ast\cdots)\bigr),$$ so that $F(k\ast f)=k\ast(D_v f)$.

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  • $\begingroup$ Thank you Carlo, I really appreciate it. I have two questions: 1. Are we garanteed that the inverse of $k$ exists even if it is numerically known? 2. I persume that you are referring to the convolution theorem when saying to calculate $q$ by Fourier transformation. I understand that if the convolution kernel is space/time invariant, the Fourier transform of $q$ is one over the Fourier transform of $k$. However, the kernel here is space/time variant. Does the theorem apply? $\endgroup$
    – Mirar
    Commented Mar 3, 2022 at 21:42
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    $\begingroup$ you are right, Fourier transformation does not simplify the calculation of the inverse kernel; what I would try is to calculate it numerically: discretize $x$ and $t$, so that $k(x,t)$ is a finite-dimensional matrix, then $q$ is the inverse matrix. $\endgroup$
    – Carlo Beenakker
    Commented Mar 3, 2022 at 22:03
  • $\begingroup$ We initially tried to calculate $f$ by constructing a matrix equation $K.vec(f)=I$, where $vec(f)$ is the vectorized form of $f$. However, we faced some difficulties. Since $f$ is a 3D data, the matrix $K$ is very large ($N_{x} \times N_{y} \times N_{z})^{2}$) and since, in our case, $k$ has a large spatial support, $K$ is not sparse and it takes a large memory space which makes it impractical to use. We even tried to downsample the data but did not help. $\endgroup$
    – Mirar
    Commented Mar 4, 2022 at 9:15
  • $\begingroup$ Please allow me to ask another related question. Is it given that $q$ exists? $\endgroup$
    – Mirar
    Commented Mar 4, 2022 at 9:22
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    $\begingroup$ well, since a non-invertible $k$ is of measure zero that should not be a problem; think of it like one big matrix; a zero eigenvalue will require fine tuning of the matrix elements, it will not happen generically. $\endgroup$
    – Carlo Beenakker
    Commented Mar 4, 2022 at 10:47

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