matrix inner product between positive semidefinite matrix and positive definite matrix 
Let $F_0, F_1, \ldots, F_m$ be a $n \times n$ symmetric matrices. We define $$F(x) := F_0 + x_1 F_1 + \cdots + x_m F_m$$ Show that if there does not exist $x \in \Bbb R^m$ such that $F(x)$ is positive definite, then there does exist a positive semidefinite matrix $H \neq 0$ such that $$\sup_x \mbox{tr} \left( F(x) H \right) \leq 0$$


Can you solve it? Contraposition may be useful, but I have no idea.
 A: Recall that $\langle A,B \rangle = \operatorname{tr}(AB)$ defines an inner product over the set of symmetric matrices.
The set $S_1 = \{F(x):x \in \Bbb R^n\}$ and the set $S_2$ of positive definite matrices are both convex, and we are given that these two sets are disjoint. By the hyperplane separation theorem, there exists a non-zero matrix $H$ and a constant $c$ such that we have $\langle X,H \rangle \leq c \leq \langle Y, H\rangle$ for all $X \in S_1$ and $Y \in S_2$.
Because $0$ lies in the closure of $S_2$, we have $c \leq \langle 0,H \rangle = 0$.
Claim: $\langle Y,H \rangle \geq 0$ holds for all positive definite $Y$.
Proof of Claim: Suppose that this does not hold. Then, there exists a positive definite $Y$ for which $\langle Y,H \rangle < 0$. For any positive $k$, we note that $kY$ is also positive definite, and $\langle kY,H \rangle = k\langle Y,K \rangle$. This means that $\inf_{Y \in S_2} \langle Y,H \rangle = -\infty$, which contradicts our earlier statement that we always have $\langle Y,H \rangle \geq c$ for some (finite) constant $c$. $\square$
Because $H$ is such that $\langle Y,H \rangle \geq 0$ holds for all positive definite $Y$, it must hold that $H$ is positive semidefinite. Thus, there is indeed a non-zero positive semidefinite matrix $H$ for which $\langle X, H \rangle \leq c \leq 0$ for all $X \in S_1$, which was what we wanted.
