I have the following equation:
$$\lambda=\displaystyle\int_{a}^b f(x)g(x)dx$$
Where $\lambda$ is a constant and I know the expresión for f(x). Is there any way of extracting the fucntion g(x)?
I have been looking at Fredholm equations, but you need 2 varibles in the kernel to be ble to solve it at least numerically.
Thanks
There is no unique function $g$ which satisfies the equation. For example, if $f(x) = 0$, then the equation has either no solutions (if $\lambda \neq 0$) or any function $g$ is a solution (if $\lambda = 0$).
This non-uniqueness is not only an abnormal behaviour at $f=0$, either. Take, for example, $f(x) = 1$ on $[0, 1]$, and take $\lambda = \frac12$. Then, the functions $g_1(x) = x$ and $g_2(x) = 1-x$ both solve the equation. More than that, for any $\mu\in [0,1]$, the function $g(x) = \mu g_1(x) + (1-\mu) g_2(x)$ also solves the equation, and so does a large number of other functions.
In fact, take any non-zero function $f$ and any function $h$ such that $$I =\int_a^b f(x) h(x) \neq 0$$ Note that such a function always exists if $f$ is a continuous non-zero function (since you can take $h=f$).
Then, for any $\lambda\in\mathbb R$, you can find a solution to the equation $$\lambda = \int_a^b f(x)g(x) dx$$
because the function $g(x) = \frac{\lambda}{I} h(x)$ satisfies the equation:
$$\int_a^b f(x) g(x) dx = \int_a^b f(x) \frac{\lambda}{I} h(x) dx = \frac\lambda I \int_a^b f(x)h(x)dx = \frac\lambda I I = \lambda$$