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Suppose I have a linear operator $K: L^2(D) \to L^2(D)$, where $D$ is some domain in $\mathbb{R}^2$ or $\mathbb{R}^3$, given by

$$\psi(x) \to K[\psi](x) = \int_D J(x - y)\psi(y) dy,$$

for $x \in D$. How would I take the gradient of the inverse of this operator, that is $\nabla_x K^{-1}[\psi](x)$?

For instance, for a regular function $f$ I would say that

$$\nabla f^{-1}(x) = (f^{-1})' \nabla f(x)= (-1)f^{-2}(x) \nabla f(x),$$

...so what can be said about the gradient of the inverse of the operator $K$?

Is it simply $\nabla K^{-1}[\psi](x) = (-1)K^{-2}[\psi](x) \nabla K[\psi](x)$?

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    I'd start by checking that $K$ is indeed invertible. And what do you mean by gradient? Gradient of the operator, or the gradient of the value of this operator? – 2017-02-10
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    @TZakrevskiy I'd say this operator is essentially a convolution, if it inverse exists it is a convolution too, and $f \mapsto \nabla f$ is a convolution operator, finally the convolution is commutative. – 2017-02-10
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    @TZakrevskiy I know that this operator is indeed invertible. By gradient, I mean in the sense that we can write $\nabla_x K[\psi](x) = \int_D J(x - y)\psi(y) dy = \int_D \nabla_x J(x - y)\psi(y) dy$ – 2017-02-10
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    @user1952009 I'm not too concerned with an explicit representation of the gradient of the inverse of the operator $K$, I'm more interested in an expression for the gradient of the inverse of $K$ like we have for some regular function $f$? – 2017-02-10

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