I see. I was, indeed, being dense. You make an interesting point. Thank you.
Ultimately, I am now uninterested in the problem since I realized I cannot solve for a unique K. If you consider the discrete case of my problem where f and g are length-N vectors, there are infinitely many...
I still can't see why defining h'(\mathbf{v}) = h(\mathbf{v}) + \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}} implies \int h'(\mathbf{v}) d \mathbf{v} = 0 . I am probably just being dense.
Instead of getting caught up in the details... Is it accurate to say you are showing me...
It seems like you are making an argument that there are infinitely many K(\mathbf{w},\mathbf{v}) so one cannot solve for a unique kernel. This is a conclusion I have already come to. However, I do not understand your argument. How can you say h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int...
Consider,
f(\mathbf{w}) = \int K(\mathbf{w,\mathbf{v}}) g(\mathbf{v}) d\mathbf{v}
where \mathbf{v},\mathbf{w} \in \mathbb{R^3}.
Is it possible to solve for the integral kernel, K(\mathbf{w,\mathbf{v}}) , if f(\mathbf{w}) and g(\mathbf{v}) , are known scalar functions and we require...