Why eigenfunction cannot be identically zero in this question? 
Below is part of the solution:

 A: Eigenvectors and eigenfunctions are precluded from being (identically) $0$ by definition.  Otherwise we would have, for any linear operator $L$ and any scalar value $r$,
$L(0) = 0 = r(0), \tag 1$
which says that every scalar is an eigenvalue of every operator.  Useless . . . 
A: An eigenvector $v$ with corresponding eigenvalue $\lambda$ for a linear operator $L$ is a nonzero vector such that the following holds:
$$Lv=\lambda v$$
That is to say that the operator $L$ acts on $v$ in the same way as multiplication by a scalar does.
Eigenfunctions are exactly the same as eigenvectors, the only difference being the name eigenfunction implies that we are talking specifically about a function space as opposed to an arbitrary vector space.  Function spaces are vector spaces, just specifically a vector space whose vectors are themselves functions.
We require that eigenvectors be nonzero because the zero vector trivially satisfies $L0=\lambda 0$ and this is incredibly uninteresting.  The interesting properties and theories surrounding eigenvectors and eigenvalues are those situations where there is a corresponding eigenspace (a linear subspace of our vector space spanned entirely by eigenvectors for that corresponding eigenvalue) of dimension strictly greater than zero.  As such, we opt to not call the zero vector an eigenvector as its existence alone as a solution to $Lv=\lambda v$ does not imply anything about the dimension of the eigenspace for $\lambda$.
