Lazarsfeld-Mukai bundles are stable on a K3 surface of picard number 1 Let $S$ be a K3 surface with $Pic(S) = \mathbb{Z}.[C]$, where $C$ is a smooth curve. Let $A$ be a complete, base point free $g_d^r$ on $C$. How to show that the Lazarsfeld-Mukai bundle $E_{C,A}$ associated to $(C,A)$ is $C$-stable on $S$?  (The bundle $E_{C,A}$ is as defined in Lazarsfeld's paper 'Brill-Noether-Petri without degenerations').
 A: Since the Lazarsfeld-Mukai bundle $E = E_{C,A}$ fits into the exact sequence
$$0 \to H^0(C,A)^\vee \otimes \mathcal{O}_S \to E \xrightarrow{\phi} K_C(-A) \to 0,$$
the first Chern class of $E$ is  $[C]$. Let $F$ be a subsheaf of $E$ and let $K = \ker(\phi_{|F})$ and $L= \mathrm{im}(\phi_{|F})$.  Since $K$ and $L$ are subsheaves of $V :=H^0(C,A)^\vee \otimes \mathcal{O}_S$ and $K_C(-A)$ respectively, one has
$$c_1(F) \cdot [C] = (c_1(K) +c_1(L))\cdot [C] \le [C]\cdot [C] = c_1(E) \cdot [C].$$
The equality holds when $L = K_C(-A)$ and $K$ is a direct sum of copies of $\mathcal{O}_S$. In this case, the quotient $E/F \simeq V/K$ is also a  direct sum of copies  of $\mathcal{O}_S$. As $H^0(E^\vee) = 0$, we conclude that $E = F$.
Therefore if $F$ is a proper saturated subsheaf of $E$ such that $\mu_C(F) \ge \mu_C(E)$, we would have
$$[C]\cdot [C] > c_1(F)\cdot [C] \ge \frac{rkF}{rkE}[C]\cdot [C] > 0,$$
which is in contradiction to the assumption that $Pic(S) = \mathbf{Z} \cdot[C]$.
(EDIT: Last inequality corrected following abx's comment.)
A: You can argue as follows. Put $E:=E_{C,A}$. You need only to know : $c_1(E)$ is the positive generator of $\mathrm{Pic}(S)$,  there is a homomorphism $\mathscr{O}_S^r\rightarrow E$ which is generically surjective, and $H^0(S,E^*)=0$.
If $E$ is not stable,  it admits a torsion-free quotient $\mathscr{F}$ with $c_1(\mathscr{F})\leq 0$.  The bidual $F$ of $\mathscr{F}$ is a vector bundle with the same $c_1$, and a generically surjective homomorphism $\mathscr{O}_S^r\rightarrow F$. Let $s:=\rm{rk}( F)$; 
 we get a generically surjective homomorphism $\wedge^s\mathscr{O}_S^r\rightarrow \det(F)$. Therefore $\det(F)=\mathscr{O}_S$, and there is a direct factor $\mathscr{O}_S^s$ of $\mathscr{O}_S^r$ such that the induced homomorphism $\mathscr{O}_S^s\rightarrow F$ has nonzero determinant, hence is an isomorphism. But this contradicts 
$H^0(S,E^*)=0$.
