Canonical module Let $(R,m)$ be a Cohen-Macaulay ring and $M$ a maximal Cohen-Macaulay $R$-module. 
Assume there exists canonical module $K_R$. Set $L:=\{f(x)|x\in M, f\in Hom_R(M,K_R)\}$. Then $M$ is $R$-module. 
I want to find some conditions are equivalent to (or implies) the following condition
$$L\subseteq mK_R$$
Please give me some examples. 
 A: It seems to me that what you would like can be rephrased the following way:
Consider the short exact sequence
$$
0\to mK_R \to K_R \to K_R/mK_R \to 0,
$$
and apply the functor $Hom_R(M,\_\_)$ to obtain the exact sequence
$$
0\to Hom_R(M,mK_R) \to Hom_R(M,K_R) \to Hom_R(M,K_R/mK_R) \to\\ 
\to Ext^1_R(M,mK_R) \to Ext^1_R(M,K_R)\to  Ext^1_R(M,K_R/mK_R)\to \dots
$$
Now, what you are asking for is that the first non-trivial morphism in this sequence,
$$
Hom_R(M,mK_R) \to Hom_R(M,K_R)
$$
is surjective (i.e., an isomorphism).
So, the actual condition is that the next morphism is zero
$$
Hom_R(M,K_R) \to Hom_R(M,K_R/mK_R),
$$
in other words that no morphism $M\to K_R/mK_R$ can be lifted to a morphism $M\to K_R$. 
An easy way this could happen would be if the latter group were zero, i.e., if $Hom_R(M,K_R/mK_R)=0$, but I don't think this can happen; $K_R/mK_R$ is a $k=R/m$-vector space. So is $M/mM$ and by Nakayama's lemma both of these are non-trivial and hence there are non-trivial morphisms between them. 
Another, somewhat more drastic way this happens is if the former group is zero, i.e., if $Hom_R(M,K_R)=0$. This actually can happen, but it probably defeats the point of your question as in that case also $L=0$ and the condition holds trivially. 
I know this is of little help, but my feeling is that this shows that it is unlikely that what you are asking for happens in a natural way under non-trivial circumstances. 
A: The case when $R$ is Gorenstein is simple: $K_R=R$, and $L\subseteq m$ iff there is no surjection $M\to R$ iff $M$ has no summand isomorphic to R. In the example you are interested in, mentioned in your comment to Sándor's answer, $M$ is indecomposable and non-free, so it holds. 
