Continuity of max of linear function on compact convex set Let $z \in R^n$ and $C \subset R^n$ be a compact convex subset. I think $$f(z) = \max_{y \in C} y^T z$$ as a function of $z$ is continuous, but don't know how to prove, or disprove, it.
 A: This is actually a special case of Danskin's Theorem. Let us look at a more general case (not the most general one). 
Suppose $f: X \times Y \to \mathbb R$ is continuous function with $X \subset \mathbb R^n$ open and $Y$ is a compact subset of some topological space $Z$. Then $ g(x) = \max_{y \in Y} f(x, y)$ is continuous. 
To prove the statement, take any $x_0 \in X$, and let $\{x_j\}_1^{\infty}$ be a sequence converging to $x_0$. We pick another sequence $\{y_j\}_1^{\infty}$ in $Y$ such that $g(x_j) = f(x_j, y_j)$. Since $Y$ is compact, we may assume $y_j \to y_0$ (possibly need to take subsequence). We note for each $y \in Y$, $f(x_j, y_j) \ge f(x_j, y)$ by the very definition of $g$. Then for every $y$,
\begin{align*}
\lim_{j \to \infty} g(x_j) = \lim_{j \to \infty} f(x_j, y_j) = f(x_0, y_0) \ge \lim_{j \to \infty} f(x_j, y) = f(x_0, y).
\end{align*}
This implies $g(x_0) = f(x_0, y_0)$. Consequently, we have
\begin{align*}
g(x_0) = f(x_0, y_0) = \lim_{j \to \infty} g(x_k),
\end{align*}
for every convergent sequence. So $g$ is continuous.
