Boundedness of the operator that multiplicates maps

Question

Let $C[0,1]$ and $C(\mathbb{R})$ 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 $\phi\in C([0,1])$ and let $T:C([0,1])\rightarrow C([0,1])$ with $T(\psi)(x)=\phi(x)\psi(x)$ and the supremum norm.
1. What is $||T||$?
2. For $\psi\in C_0(\mathbb{R})$, why do we have $||\psi||<\infty$?

What I know:
1. So for the norm we know that $$||T||=\sup_{\psi\in C([0,1])}\{||T\psi||=\sup_{x\in[0,1]}|\phi(x)\psi(x)|:\sup_{x\in[0,1]}|\psi(x)|\leq 1\}.$$

2. We want to see that $\sup_{x\in\mathbb{R}}|\psi(x)|<\infty$.

Answer 1

2. Since $\psi$ is continuous, it can not diverge when $x\to x_0$ for some $x_0\in\mathbb R$. It can not even diverge at $\pm\infty$ by definition. Thus $\psi$ is bounded.

Answer 2

1. $|T(\psi)(x)|=|\phi(x)|*|\psi(x)| \le ||\phi||*||\psi||$ for all $ x \in [0,1]$

Therefore $||T(\psi)|| \le ||\phi||*||\psi||$ , thus $||T|| \le ||\phi||$.

Take $\psi(x)=1$ to get $||T|| = ||\phi||$.

Answer 3

To have 2. spelled out more concretely:

$\lim_{x\to\infty} \psi(x)=0$ means that for all $\epsilon>0$ there exists an $r$ so that for all $x>r$ you have $|\psi(x)|<\epsilon$. Analogously for $\lim_{x\to-\infty} \psi(x)=0$ you get for every $\epsilon$ an $r'$ so that $x<r'$ gives you $|\psi(x)|<\epsilon$.

If you take $\epsilon=1$ for example you then find that outside of $[r',r]$ the function $\psi$ has absolute value smaller than $1$. So it is bounded outside of $[r',r]$. But $[r',r]$ is compact and $\psi$ is continuous, so $|\psi|$ is also bounded on $[r',r]$. Take the maximum of the two bounds to get a bound on all of $\mathbb R$.