I am recently feeling headache on a problem about eigenfunctions. Here it is: suppose we have an eigenvalue equation as following $$\hat{H} \Psi(x) = \lambda \Psi(x)$$ Assume that $x$ is discrete and $\hat{H}$ is a finite dimensional differential operator, I deduce that we should have finite number of eigenvalues and corresponding eigenfunctions.
Let $\phi_1(x)$ and $\phi_2(x)$ be two different eigenfunctions corresponding to two distinct eigenvalues $\lambda_1$ and $\lambda_2$. We can construct a new function $\phi(x)=A\phi_1(x)\phi_2(x)$, where $A$ is a constant. We can always derive that $$\hat{H} \phi(x) = (\lambda_1 + \lambda_2) \phi(x)$$ Given that $\hat{H}$ is a linear differential operator. In this way we can construct infinite number of eigenfunctions, which is conflict from my previous conclusion that the number of eigenvalues and eigenfunctions should be finite. On the other hand, I have learn that the eigenvector corresponds to distinct eigenvalue are always linearly independent. So $\phi(x)$ is a new eigenfunction different from $\phi_1(x)$ and $\phi_2(x)$.
Of course I was wrong at some point, but I cannot figure out where, anyone help me get out of this problem. Thank you in advance.