Show sequence equicontinuous I don't know how to prove this question:
Let $X$ be a compact space, and let $(T_{n})$ be a sequence of positive linear operators on $C(X)$. Also, let $f \in C(X)$ be a strictly positive function. Then if $T_{n}(f) \rightarrow f$ (uniformly), then the sequence $(T_{n})$ is equicontinuous. 
Now since this is a strictly positive function, I am aware that the function $f(x) \geq 0$ for all $x \in X$ but how do I goes about in proving the next part of the sentence? 
Hope to have someone to shed some light on this. Thank You.
 A: I'll only deal with real-valued functions, but the complex case is easy to get from this.
A positive operator on $C(X)$ is continuous because its norm satisfies $\|T\| = \|T(1_X)\|_{\infty}$:
Writing $1_X$ for the constant function $x \mapsto 1$, we have for all $g \in C(X)$ that
$$
-\|g\|_\infty \cdot 1_X \leq g \leq \|g\|_\infty \cdot 1_X,
$$
so positivity of $T$ yields
$$
-\|g\|_\infty T(1_X) \leq T(g) \leq \|g\|_\infty T(1_X).
$$
and therefore $\|T(g)\|_\infty \leq \|g\|_\infty \cdot \|T(1_X)\|_\infty$. This shows that $\|T\| \leq \|T(1_X)\|_\infty$ and taking $g = 1_X$ we see that a positive operator must have $\|T\| = \|T(1_X)\|_\infty$.
On the other hand, we are given that $\|T_n(f) - f\|_\infty \to 0$, where $f$ is assumed to be strictly positive. By compactness of $X$ and continuity of $f$ there is $x_0 \in X$ such that $0 \lt f(x_0) \leq f(x)$ for all $x \in X$. This tells us that $f(x_0) 1_X \leq f$ and by positivity of the operators $T_n$ we conclude that $0 \leq f(x_0) T_n(1_X) \leq T_n(f)$ whence $\|T_n\| = \|T_n(1_X)\|_\infty \leq \frac{1}{f(x_0)} \|T_n(f)\|_\infty$.
However, since $\|T_nf - f\|_\infty \to 0$, there exists a constant $C$ such that $\|T_n(f)\|_\infty \leq C$ for all $n$. In conclusion, $\|T_n\| \leq \frac{C}{f(x_0)}$ for all $n$, which is equicontinuity of the family of linear operators $T_n$.
