# Norm of an multiplication operator

Let $C(\mathbb{R})$ be the vector spaces of all continuous maps on $\mathbb{R}$. Then let $C_0(\mathbb{R})\subset C(\mathbb{R})$ with all maps $g(x)$ that go to $0$ as $x\rightarrow\pm\infty$.

Fix a $f\in C(\mathbb{R})$ and define $M_f:C(\mathbb{R})\rightarrow C(\mathbb{R})$ with $M_f(g)=g(x)*f(x)$. Suppose that $f$ is such that $M_f(C_0(\mathbb{R}))\subset C_0(\mathbb{R})$ and $M_f|_{C_0(\mathbb{R})}$ is bounded.
1. Why is f bounded?
2. Compute $||M_f||$?

To 2. I know that if the operator would just be defined on $[0,1]$ we have that $||M_f||=||f||_\inf$. Isn't it the same for this operator as well?

For $x \in \def\R{\mathbf R}$, $\def\eps{\varepsilon}\eps > 0$, let $g_{x,\eps}$ denote the function $$g_{x,\eps}(t) = \begin{cases} 0, & t \le x-\eps\\ \frac{t-x+\eps}{\eps}, & x-\eps \le t\le x\\ \frac{x+\eps-t}{\eps}, & x \le t \le x+\eps\\ 0, & t \ge x + \eps \end{cases}$$ Then $g_{x,\eps} \in C_0(\R)$ has support in $[x-\eps, x+\eps]$ and $\def\norm#1{\left\|#1\right\|}\norm{g_{x,\eps}}_\infty = 1$. As $M_f$ is bounded on $C_0(\R)$, we have $$\def\abs#1{\left|#1\right|}\abs{f(x)} = \abs{f(x)g_{x,\eps}(x)} \le \norm{fg_{x,\eps}}_\infty = \norm{M_fg_{x,\eps}}\le \norm{M_f}$$ Hence, $f$ is bounded with $\norm{f}_\infty \le \norm{M_f}$. On the other hand, we have $$\norm{M_fg}_\infty = \norm{fg}_\infty\le \norm f_\infty \norm g_\infty$$ giving $\norm{M_f}\le \norm f_\infty$. That is $\norm{M_f} = \norm f_\infty$.