Let $C(\mathbb{R})$ be the vector spaces of all continuous maps on $[0,1]$ and $\mathbb{R}$ respectively. Then let $C_0(\mathbb{R})\subset C(\mathbb{R})$ with all maps $\psi(x)$ that go to $0$ as $x\rightarrow\pm\infty$.
Fix a $\phi\in C(\mathbb{R})$ and define $T:C(\mathbb{R})\rightarrow C(\mathbb{R})$ with $T(\psi)(x)=\phi(x)\psi(x)$ and the supremum norm. Suppose that $\phi$ is such that $T(C_0(\mathbb{R}))\subset C_0(\mathbb{R})$ and $T|_{C_0(\mathbb{R})}$ is bounded.
1. Why must $\|\phi\|<\infty$?
2. What is $\|T\|$?
What I know:
The supremum norm for $f\in C_0(\mathbb{R})$ is $$\|f\|:=\sup_{x\in\mathbb{R}}\{|f(x)|\}$$
and the norm of an operator is $$\|T\|:=\sup_{\|f\|\leq1}\{\|Tf\|\}.$$
So we know that for all $\psi\in C_0(\mathbb{R})$ we get $\phi\psi\in C_0(\mathbb{R})$ and $$\|T\lvert_{C_0(\mathbb{R})}\|=\sup_{\|\psi\|\leq1,\psi\in C_0(\mathbb{R})}\{\|T\psi\|\}=\sup_{\|\psi\|\leq1,\psi\in C_0(\mathbb{R})}\{\|\phi\psi\|\}<\infty.$$
How do 1. and 2. follow?
Suppose $\phi$ is not bounded. Then for every $n$ you have an $x_n$ so that $|\phi(x_n)|>n$. Choose some bump function $\psi_n$ in $C_0(\mathbb R)$ that is $1$ at $x_n$ and has norm $\|\psi_n\|=1$, then $|T(\psi_n)(x_n)|=|\phi(x_n)\psi(x_n)|≥n$. So also $\|T(\psi_n)\|≥n=n\|\psi_n\|$, this is a contradiction to there existing a $C$ so that $\|T(\psi)\|≤C\|\psi\|$ for all $\psi\in C_0(\mathbb R)$. So you gotta have that $\phi$ is bounded.
Now let $\|\phi\|<\infty$, you have $\|\phi\cdot\psi\|≤\|\phi\|\cdot\|\psi\|$ for all bounded functions $\psi$, so $\|T\|≤\|\phi\|$.
On the other hand let $x_n$ be a sequence so that $|\phi(x_n)|\to\|\phi\|$. Taking bump functions $\psi_n$ as before gives that $\|T(\psi_n)\|≥|\phi(x_n)\psi_n(x_n)|=|\phi(x_n)|$, so $\|T\|≥|\phi(x_n)|$ for all $n$ and thus also $\|T\|≥\|\phi\|$ by taking limits.
These two together give $\|T\|=\|\phi\|$.