Integral equation with bounded unknown kernel

[JohnXYZ]

I need to solve an integral equation of the form

$$\forall \omega \in [0,1], ~ \int_{\mathbb{R}} K(\omega,y)f(y)dy = \omega$$

where

- f is known and positive with $$\int_{\mathbb{R}} f(y)dy = 1$$

- K: [0,1] x R -> [0,1] is the unknown kernel

I am looking for a solution other than K(omega,y) = omega. I do not know if such a solution exists, so I am looking either for a solution, or for a proof that K(omega,y) = \omega is the only solution.

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Let me tell you more about my attempts :

- looking for K(omega,y) = a(omega)b(y) fails because it leads to a(omega) = \omega and b(y) = 1

- looking for K(omega,y) = a(omega-y) and trying to solve the convolution equation $$\forall \omega \in [0,1], ~ (a \star f)(\omega) = \omega$$ fails (I proved it with Fourier transforms, although I'm not 100% confident about my proof)

- setting K(omega,y) = omega + H(omega,y) and trying to solve $$\int_{\mathbb{R}} H(\omega,y) f(y) = 0$$ with either H(omega,y) = a(omega)b(y) or H(omega,y) = a(y-omega) fails as well

There may not be a solution but I hope you can help me find one or prove that it's impossible ! Thank you

 

[Stephen Tashi]

One way of thinking about it is to assume we can differentiate inside the integral sign with respect to $\omega$

$$\int_{\mathbb{R}} K'(\omega,y)f(y)dy = 1$$
$$\int_{\mathbb{R}} K"(\omega,y)f(y)dy = 0$$

We can consider $K"$ to be a member of a family of functions of $y$ parameterized by $\omega$ such that each member of the family is orthogonal to $f(y)$ on $\mathbb{R}$.

Suppose we have an orthogonal basis $\{f_1(y), f_2(y),f_3(y)...\}$ for some space of functions and that $f = f_1$ Then $\omega f_2 + (1 - \omega) f_3$ is a family of functions that is orthogonal to $f$.

I'm just making these suggestion off the top of my head. I haven't tried to work this out with any concrete example.

 

[JohnXYZ]

Thank you for your answer, I will also try your approach. Working from K'' may still be difficult since we have a boundedness condition on K.