Eigenstates of Ladder Operators According ot Griffith's Intro to Quantum Mechanics (page 147), if some function $f$ is an eigenfunction of $L^{2}$, then $L_{-}f$ is also an eigenfunction of $L^{2}$.  
Is $f$ also an eigenfunction of of $L_{-}$?  In general, what are the eigenfunctions of $L_{-}$ and $L_{+}$?
 A: 
Is $f$ also an eigenfunction of $L_-$?

In general, no. Introduce the notation
$$
L^2 |\ell;m\rangle=\ell(\ell+1)|\ell;m\rangle \tag{1}
$$
and
$$
L_z|\ell;m\rangle=m |\ell;m\rangle \tag{2}
$$
In general$^1$,
$$
L_\pm|\ell;m\rangle=\sqrt{\ell(\ell+1)-m(m\pm1)}\;|\ell;m\pm 1\rangle \tag{3}
$$
which means $|\ell;m\rangle$ is not an eigenvector of $L_\pm$, because when acting with $L_\pm$ on $|\ell;m\rangle$, you dont get a scalar multiple of $|\ell;m\rangle$, but a different vector.
In the specific case $|\ell;\ell\rangle$ (i.e., when $m=\ell$), we have
$$
L_+|\ell;\ell\rangle=0 \tag{4}
$$
which means $|\ell;\ell\rangle$ is an eigenvector of $L_+$, with eigenvalue 0. We can say the same about $L_-$: $|\ell;-\ell\rangle$ is an eigenvector of $L_-$, with eigenvalue $0$.

In general, what are the eigenfunctions of $L_\pm$?

To begin with, $L_\pm$ are not hermitian, so there is no guarantee these are diagonalisable, and if they are, the eigenvalues won't be real (i.e., $L_\pm$ is not an observable). In the paragraph above, we argued that $|\ell;\ell\rangle$ is an eigenvector of $L_+$, so that at least there exist one eigenvector. Now we prove that $|\ell;\ell\rangle$ is the only eigenvector of $L_+$.
Let us suppose that there exist a set of vectors $|\alpha;\ell\rangle$ such that
$$
L_+|\alpha;\ell\rangle=\alpha|\alpha;\ell\rangle \tag{5}
$$
and
$$
L^2|\alpha;\ell\rangle=\ell(\ell+1)|\alpha;\ell\rangle \tag{6}
$$
(note that we can do this because $[L_+,L^2]=0$).
As the set $\{|\ell;m\rangle\}$ is a basis, we can write $|\alpha;\ell\rangle$ as a linear combination of these vectors:
$$
|\alpha;\ell\rangle=\sum_{m=-\ell}^{+\ell} c^m_\ell|\ell;m\rangle \tag{7}
$$
It's fairly easy to check that $(5)$ can only be satisfied if all the coefficients $c^m_\ell=0$ except for $c^\ell_\ell$, so that
$$
|\alpha;\ell\rangle=|\ell;\ell\rangle \tag{8}
$$
easily follows.
In the case of the simple harmonic oscillator, where the algebra is similar, the situation is different: there, we have an expression similar to $(7)$, but where the sum is over $n=0,1,\cdots,\infty$. In this case, the conclusion is different, because there is and infinite number of terms in the sum. Now, its easy to prove that there exist a set of non-zero coefficients $c_n$, which means that there exist non-zero eigenvectors of the rising/lowering operators. This are called coherent states, and are quite fun.

$^1$ The proof of this expression is pretty standard, and can be found online and in any book on QM. I'm going to reproduce the proof here to make the post more self-contained.
The algebra of the angular momentum operators is
$$
[L_x,L_y]=iL_z\qquad [L_y,L_z]=iL_x \qquad [L_z,L_x]=iL_y \tag{9}
$$
If we define $L_\pm=L_x\pm i L_y$, then its easy to check that $(9)$ is equivalent to
$$
[L_z,L_\pm]=\pm L_\pm \tag{10}
$$
For example, $[L_z,L_+]=[L_z,L_x+iL_y]=[L_z,L_x]+i[L_z,Ly]$ which by virtue of $(9)$ equals $=iL_y+L_x=L_+$.
With this, we can prove $(3)$. Let
$$
|\varphi\rangle\equiv L_+|\ell;m\rangle \tag{11}
$$
by definition. If we act on the left with $L_z$, we get
$$
L_z|\varphi\rangle=L_z L_+|\ell;m\rangle \tag{12}
$$
Next, write $L_zL_+=L_+L_z+[L_z,L_+]$ (this should be rather obviously true: just expand the commutator, and check that it works). As we know that $[L_z,L_+]=L_+$, we get
$$
(12)=(L_+ L_z+L_+)|\ell;m\rangle \tag{13}
$$
which, using $L_z|\ell;m\rangle=m|\ell;m\rangle$, equals
$$
(12)=(1+m)L_+|\ell;m\rangle \tag{14}
$$
Finally, note that $L_+|\ell;m\rangle$ is, by definition, $|\varphi\rangle$, which means that
$$
L_z|\varphi\rangle=(m+1)|\varphi\rangle \tag{15}
$$
This relation is very important! Try to think about it for a minute. Look at it carefully. This relation means that $|\varphi\rangle$ is an eigenvector of $L_z$, and its eigenvalue is $m+1$. Therefore, we must have $|\varphi\rangle\propto |\ell;m+1\rangle$, for $|\ell;m+1\rangle$ is defined as the eigenvector of $L_z$ with eigenvaule $m+1$.
Therefore, we can write
$$
L_+|\ell;m\rangle=c|\ell;m+1\rangle \tag{16}
$$
where $c$ is a normalisation constant, which is easy to find, because we know that $L_- L_+=L^2-L_z^2-L_z$:
$$
|c|^2=\langle\ell;m|L_-L_+|\ell;m\rangle=\langle\ell;m|L^2-L_z^2-L_z|\ell;m\rangle=\ell(\ell+1)-m^2-m \tag{17}
$$
where I used $L^2|\ell;m\rangle=\ell(\ell+1)|\ell;m\rangle$ and $L_z|\ell;m\rangle=m|\ell;m\rangle$. This completes the proof of $(3)$.
A: This isn't too difficult to figure out, construct the L+/- matrices corresponding to a certain value of l, and then you can easily find their eigenvectors using only matrix algebra. 
A: Actually when you define  ladder operators $L_+$ and $L_-$  like
$$
\begin{aligned}
L_+&= L_x+iL_y\\
L_-&= L_x-iL_y 
\end{aligned}
$$
and $L^2$ commutes with  $L_x$ and $L_y$.
Therefore, the commutator of  $L^2$  with $L_-$ and $L_+$ vanishes;
$$
L_+ Y(l,m)=\hbar \sqrt{l(l+1)-m(m+1)}Y(l,m+1)
$$
and similarly for $L_-$, where $Y$'s are the spherical harmonics- a complete set of angular functions which are eigen functions of $L^2$ and $L_z$  and other commuting operators with $L^2$ and $L_z$.
$$[L_+, L_z] = - \hbar L_+$$
and similarly for $L_-$
$L_+$ raises the $m$ value of $Y$'s by one unit of $\hbar$ and $L_-$ lowers the  $m$ value of $Y$'s by one unit.
The index $m$ can take values from $-l$ to $+l$ with interval of 1, so 
$$L_+Y(l,l) = 0$$ and $$L_- Y(l,-l) =0$$
for $l = 0,1,2,3,\dots$; $m$ can take values $-l,-l+1,\dots,0,1,2,\dots,l-1,+l$
I think the above can clarify the picture of eigenfunctions.
Ref. http://quantummechanics.ucsd.edu/ph130a/130_notes/node209.html
