My issue is this: Let's say I have the operator $\hat{X} = \frac{d}{dx}$ and I want to find complex-valued functions $f:\mathbb{C}\rightarrow\mathbb{C}$ which satisfy $\hat{X} f = \lambda f,$ with $\lambda \in \mathbb{C} $, meaning I am looking for eigenfunctions of $\hat{X}$
Now I want to test if the functions $f_1(x) = e^{-a\mathrm{i}x^2}$ and $f_2(x)=e^{bix}$ satisfy the condition. When I apply the operator, it gives me
$$\hat{X}f_1(x) = \frac{d}{dx}e^{-a\mathrm{i}x^2} = -2a\mathrm{i}x\cdot e^{-a\mathrm{i}x^2} = -2a\mathrm{i}x \cdot f_1(x)$$ and $$\hat{X}f_2(x) = \frac{d}{dx}e^{b\mathrm{i}x} = b\mathrm{i}\cdot e^{b\mathrm{i}x} = b\mathrm{i} \cdot f_2(x)$$
At first glance, I would say both functions are eigenfunctions of $\hat{X}$ with the respective eigenvalues of $\lambda_1 = -2a\mathrm{i}x$ and $\lambda_2=b\mathrm{i}$, but I cannot get rid of the feeling that this conclusion is wrong for $f_1(x)$. The fact that $\lambda_1$ depends on $x$ makes me uncomfortable, and I almost want to say that it is not an eigenvalue since it is essentially another function of $x$, but I don't know why or if this feeling is even warranted.
Is a function satisfying $\hat{X}f(x) = \lambda(x)f(x)$ an eigenfunction of $\hat{X}$, or does $\lambda$ need to be a fixed scalar independent of the variable $x$? , This might be a strange question, but I am unsure if I forget something about the property of eigenvalues.