Convergence of positivity preserving operators Let $X$ and $Y$ be normed vector spaces and $Z = L(C_b(X), C_b(Y))$ be the set of linear operators from $C_b(X)$ to $C_b(Y)$, where $C_b(X)$ and $C_b(Y)$ are the sets of continuous real valued bounded functions on $X$ and $Y$ respectively. 
An operator $T \in Z$ is called positivity preserving iff
$$f(x) \geq 0 \ \text{for all x} \qquad \Longrightarrow \qquad Tf(x) \geq 0 \ \text{for all} \ x \in X.$$

Let $(S_n)_{n \in \mathbb{N}}$ be a family of linear operators in $Z$
  such that $S_{n + 1} - S_n$ is positivity preserving for each $n \in
 \mathbb{N}$. Prove that $(S_n)_{n \in \mathbb{N}}$ converges in
  operator norm iff $(S_n1)_{n \in \mathbb{N}}$ converges in
  $\sup$-norm, where $1$ is the constant function $x \longmapsto 1$.

What I did so far. It was a hint in the exercise that we first should show that each positivity preserving $T \in Z$ is automatically continuous and that furthermore, $\| T \| = \| T1 \|_{\infty}$, where $1$ is again the constant function $x \longmapsto 1$. I was able to show the continuity, more exactly I was able to show that $\| T \| \leq \| T1 \|_{\infty}$. I wasn't able to show "$\geq$", and I am furthermore not sure, whether the backwards inequality is true.
Question. If $\| T \| \leq \| T1 \|_{\infty}$ is true, then I think it is more or less clear to show the "$\Longrightarrow$" in the exercise with the $(S_n)$. But how could one show the other direction? If $\| T \| = \| T1 \|_{\infty}$ holds, the other direction would be also clear but I can't see why $\| T \| \geq \| T1 \|_{\infty}$ should be true.
 A: Your hint asks you to show $\|T\|=\|T1\|_\infty$ for positive operators.
Note first that $\|1\|_\infty=1$ so that $\|T\| =\| T\|\  \|1\|_\infty ≥ \|T1\|_\infty$ by sub-multiplicativity of the operator norm. Further since $T$ is positive and $\|f\|_\infty\pm f$ is a positive function for any $f$ you have that
$$T(\|f\|_\infty\  1 \pm f)\ (x) = \|f\|_\infty T(1)\ (x) \pm T(f)\ (x)≥0$$
for all $x$ in $Y$. In particular $|T(f)\ (x)| ≤ \|f\|_\infty \ |T(1)\ (x)| ≤ \|f\|_\infty \ \|T1\|_\infty$ for all $x$, implying $\|T(f)\|_\infty ≤ \|T1\|_\infty \ \|f\|_\infty$ for all $f$, giving $\|T\|≤\|T1\|_\infty$.
Now how does this equality help? Well if $S_{n+1}-S_n$ is positive then so is
$$S_n-S_m = \sum_{k=m}^{n-1} S_{k+1}-S_k$$
for all $n>m$, since that is a sum of positive operators. In particular if $S_n1$ converges it must be Cauchy and then $\|S_n-S_m\|=\|(S_n-S_m)1\|_\infty$ is Cauchy. Since $L(C_b(X),C_b(Y))$ is complete this implies that $S_n$ converges in operator norm.
On the other hand if $S_n$ converges in operator norm you have that $S_n1$ must converge indepent of any additional assumptions.
