How can i show this Function is Coercive? Let $M_n$ denote the set of real matrices with real entries and order $n$. Let $D=\{X\in M_n:\ \det(X)>0 \}$, where $\det$ stands for determinant. Define $F:D\rightarrow\mathbb{R}$ by $$F(X)=\operatorname{tr}(X)-\lambda\ln(\det(X))$$
where $\lambda>0$ and $\operatorname{tr}(X)$ is the trace of $X$.
I want to show that $F$ is coercive, i.e. if $X_k\rightarrow \partial D$ or $\|X_k\|\rightarrow\infty$ then $\limsup F(X_k)=+\infty$.
Thanks

The claim is false as stated. Let $n = 2$, consider $X_n = \begin{pmatrix} -n & 0 \\ 0 & -n \end{pmatrix}$. Since $\det X_n = n^2 > 0$ we have that $X_n \in D$.  By positive homogeneity of norms $\|X_n\| = n \|X_1\| \nearrow \infty$. But $\ln \det X_n = 2 \ln n$, while $\operatorname{tr} X_n = -2n$, so $F(X_n) = - 2n - 2 \ln n \searrow -\infty$. 

Edit [A slightly different counterexample]: 
Again take $n = 2$. Take $X_n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$. We have that the trace of $X_n$ is always 2, while the determinant is always 1. So $F(X_n) = 2$ for all $n$. Both the Frobenius norm and the max norm on $X_n$ satisfy $\|X_n \| \geq n$ and this gives another family of counterexamples. 
Furthermore, this example gives also an illustration of why you need to be precise about $X_k \to \partial D$. Observe that the matrix
$Y_n = \begin{pmatrix} 1 & n \\ 1/n & 1\end{pmatrix}$ has vanishing determinant, and hence belongs in $\partial D$. In any norm 
$$ \|X_n - Y_n\| = \left\| \begin{pmatrix} 0 & 0 \\ 1/n & 0\end{pmatrix} \right\| \searrow 0 $$
so we in fact have not only do $X_n$ blow up in norm, it also asymptotically approaches $\partial D$! 
(On the other hand, the statement that if $X_n \to X_0\in \partial D$ we have $F(X_n) \nearrow \infty$ is true by virtue of continuity of the trace function on $M_n$.) 
