irreducible highest weight modules Let $\mathfrak{g}$ be a simple Lie algebra. Let $M_{\lambda}$ be the Verma module over $\mathfrak{g}$ of highest weight $\lambda$ and $L_{\lambda}$ be the irreducible $\mathfrak{g}$-module of highest weight $\lambda$. 
For a $\mathfrak{g}$-module $U$, the dual $\mathfrak{g}$-module $U^*$ is defined by the rule $x|_{U^*}=(-x|_U)^*$ for any $x\in \mathfrak{g}$. Let $w$ be the longest element of the Weyl group of $\mathfrak{g}$. How to show the following?
(1) If $\lambda$ be a dominant integral weight. Then the $\mathfrak{g}-module$ $L^*_{\lambda}$ dual to $L_{\lambda}$ is isomomorphic to $L_{-w(\lambda)}$.
(2) $Hom_{\mathfrak{g}}(\mathbb{C}, U\otimes L_{\lambda}) = Hom_{\mathfrak{g}}(L^*_{\lambda}, U)$?
I know that $Hom(A, B\otimes C) = Hom(Hom(A, B), C)$. But it seems that we can not use this to show (2).
Thank you very much.
 A: Your statement of the hom-tensor adjunction is incorrect, and we need to use slightly more than the adjunction to prove what you want.  In particular, we need that if $V$ is finite dimensional, then $V^**\cong V$ and $\hom(V,W)\cong V^*\otimes W$.  Using the fact that $L_{\lambda}$ is finite dimensional, we have $$\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda})\cong\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda}^{**}) \cong \hom_{\mathfrak g}(\mathbb C, \hom_{\mathbb C}(L_{\lambda}^*,U))$$
Applying the adjunction, this is isomorphic to $\hom_{\mathfrak g}(\mathbb C\otimes L_{\lambda}^*,U)\cong \hom_{\mathfrak g}(L_{\lambda}^*,U)$.
For part (1), we need a few ideas (the upshot of which are given by Jyrki).  If $V$ is a representation, I will let $V_{\lambda}$ denote the weight space of weight $\lambda$.


*

*If $V$ is a finite dimensional representation of $\mathbb g$, $\lambda$ a weight, and $w$ an element of the Weyl group, then $\dim V_{\lambda}=\dim V_{w\lambda}$.  We can see this by noting that for each root $\alpha$, we have a copy of $\mathfrak{sl}_2$ which acts on $\bigoplus_i V_{\beta+i\alpha}$ for every weight $\beta$, and standard rep theory for $\mathfrak{sl}_2$ shows that the equality holds when $w=s_{\alpha}$ is a reflection.

*If $V$ is a finite dimensional representation and $\lambda$ is a weight, then $\dim V_{\lambda}=\dim V^*_{-\lambda}$.  This can be shown by taking a basis of weight vectors.  If $v$ is a weight vector of weight $\lambda$, the corresponding dual vector $v^*$ has weight $-\lambda$.

*Irreducible finite dimensional $\mathfrak g$-reps are classified by integral dominant weights via highest weight vectors.

*If $V$ is irreducible of highest weight $\lambda$, the weights in $V$ will be the convex hull of the weights $w\lambda$, for $w$ in the Weyl group.


Thus, it suffices to show that $L_{\lambda}^*$ has highest weight $-w_0(\lambda)$ where $w_0$ is the longest length word in the Weyl group.  As Jyrki said, this follows from the fact that $w_0(\alpha)<0$ for every positive root $\alpha$, as it will be the only dominant weight of the form $-w\lambda$ for $w$ in the Weyl group.
A: For (1): You undoubtedly already know that the lowest weight of $L_\lambda$ is $w(\lambda)$: The longest element $w$ maps all the positive roots to negative roots, and hence reverses the partial order. Put this together with the fact that the weights of $V^*$ are the negatives of the weights of $V$, and that negation also reverses the partial order of the weight lattice, and you get that ...
Edit: Oh, and $L_\lambda^*$ is simple, because it's the dual of a simple f.d. module.
