Currently I am reading Behrend's $M$-Structure and Banach-Stone Theorem. Behrend introduced $M$-bounded operator at page $55.$
Let $T$ be an operator on a Banach space $X.$ $T$ is said to be M-bounded if there is a $\lambda>0$ such that for every $x\in X,$ $Tx$ is contained in every ball which contains $\{\mu x:\mu\in\mathbb{R}, |\mu|\leq \lambda\}.$
Note that $$\{\mu x:\mu\in\mathbb{R}, |\mu|\leq \lambda\} = \text{conv}\{-\lambda x,\lambda x\}$$ where $\text{conv}\{-\lambda x,\lambda x\}$ refers to the convex hull of two points $-\lambda x$ and $\lambda x.$
Let $Q$ be a locally compact Hausdorff space and $E$ be a Banach space. Behrend further introduced the multiplication operator on the function space $C_0(Q,E),$ the collection of all $E$-valued functions on $Q$ which vanish at infinity.
For each $h\in C_0(Q),$ define the multiplication operator $M_h:C_0(Q,E)\to C_0(Q,E)$ by $$M_hf(s)=h(s)f(s), \quad \text{ for all }s\in Q.$$
Then Behrend proved the following lemma:
Lemma:For each $h\in C_0(Q),$ the operator $M_h$ is $M$-bounded.
Proof: Let $\lambda = \|h\|_\infty.$ Fix $f\in C_0(Q,E).$ Pick $g\in C_0(Q,E)$ and $r>0$ such that $$\text{conv}\{-\lambda f,\lambda f\}\subseteq B(g,r)$$ where $B(g,r)=\{k\in C_0(Q,E): \|k-g\|_\infty\leq r\}.$ It is easy to verify that $$M_hf\in B(g,r).$$ Therefore, $M_h$ is $M$-bounded.
Question: How to show that $$M_hf\in B(g,r)?$$
The following is my 'unfinish' attempt:
If $M_hf\in\text{conv}\{-\lambda f,\lambda f\},$ then we are done. However, I think it is not true as $$M_hf=\frac{\|h\|-h}{2\|h\|}(-\lambda f) + \frac{\|h\|+h}{2\|h\|}(\lambda f)$$ where $h$ is not a scalar. I also tried to manipulate $$\|M_hf-g\|_\infty = \sup_{s\in Q}\|h(s)f(s)-g(s)\|,$$ but to no avail.
Any hint is appreciated.
The result holds due to the 'point-wise' nature of the supremum norm. You have that $$ \|hf - g\|_\infty = \sup_{x\in Q} \| h(x) f(x) - g(x) \|_E \leq \sup_{x\in Q} \sup_{|\lambda'|\leq \|h\|_\infty} \| \lambda' f(x) - g(x) \|_E \leq r, $$ where the last estimate holds since conv$\{-\lambda f, \lambda f\} \subset B(g,r)$ and $|\lambda'| \leq \lambda = \|h\|_\infty$. Per definition this yields that $M_h f \in B(g,r)$.
I should clarify the reasoning behind the first inequality, by request in the comment. First, we have that, $$ \sup_{x\in Q} \| h(x) f(x) - g(x) \|_E \leq \sup_{x,y\in Q} \| h(y) f(x) - g(x) \|_E, $$ which is clear, since $x,y\in Q$ includes $x = y$. Moreover, by assumption $\|h(y)\|_E \leq \|h\|_\infty$. Hence, $$ \sup_{x,y\in Q} \| h(y) f(x) - g(x) \|_E \leq \sup_{x\in Q, |\lambda'| \leq \|h\|_\infty} \| \lambda' f(x) - g(x) \|_E $$ as claimed. The original notation was slightly missleading, so I changed it above.