bounded functions, norms 
Instead of answering this question could someone possibly explain what I need to do? I don't fully understand what the question is asking, firstly $M$ has been fully defined so how do we know $|M|$? also where it says $m$ is a bounded function, does this imply we can take any bounded function? for example $\cos(x)$?
 A: For part a):
$\|M\|$ is defined via
$$\|M\|:=\sup_{f \in B(\mathbb R),\ \|f\|_\infty ≤1} \|M(f)\|_\infty$$
Then since for example $\|M(1)\|=\|m\cdot 1\|_\infty = \|m\|_\infty$ and $\|M(f)\|=\|m\cdot f\|_\infty ≤ \|m\|_\infty \|f\|_\infty$ it follows that $\|M\| = \|m\|_\infty$.
For part b):
If $m$ has no zeros on $\mathbb R$, then the pointwise inverse $m^{-1}$ of $m$ exists. Note that:
$$m^{-1}(x)\cdot (m(x)^\mathstrut \cdot f(x)) = f(x)$$
for all $x$ and all $f$. So you would suppose the inverse of $M$ to be given by the multiplication of $m^{-1}$ (which we will from now one denote by the operator $N$, in other words $(N(f))\, (x) := m^{-1}(x)\cdot f(x)$).
But this is not true, consider for example $m(x)=e^{-x^2}$, then $m^{-1}(x) = e^{x^2}$ which is not a bounded function, and application of $N$ on a bounded function need not return a bounded function. (For example $N(1)=m^{-1}$)
If $m^{-1}$ is bounded, then $N$ is once again well defined as a linear map $B(\mathbb R) \to B(\mathbb R)$, and $N$ is the inverse of $M$. $m^{-1}$ is bounded iff there exists a $c>0$ so that $0<c<|m(x)|$ for all $x \in \mathbb R$, that is if $m$ is bounded from below.
