Show that this function is uniformly continuous on $\mathbb{R}^d$

Question

Let $f\in C_b(\mathbb{R})$ ($C_b(\mathbb{R})$ denotes the set of real-valued continuous bounded functions on $\mathbb{R})$ and $I_n:=[-n,n]$ be a compact interval. In my notes it is written:

Let $u$ be the function in $C_b(\mathbb{R})$ which equals $1$ on $I_{n}$, vanishes off $I_{n+1}$ and is affine on $[n,n+1]$ and on $[-n-1,-n]$. If we set $f':=u\cdot f$, then both functions $u$ and $f'$ are uniformly continuous on $I_{n}$ and vanish identically in $\mathbb{R}\setminus I_{n}$, hence they are uniformly continuous on $\mathbb{R}$.

It also says that the same approach can be used for $f\in C_b(\mathbb{R}^d)$ by replacing $I_n$ with $\{x\in\mathbb{R}^d:\lvert x\lvert\leq n\}$ and the function $u$ with the radial function $x \mapsto u(\lvert x\lvert)$.

Questions:

1. Shouldn't $n$ be redefined as $n+1$ for the functions to vanish in $\mathbb{R}\setminus I_{n}$?.

2. I don't see how the argument generalizes to $\mathbb{R}^d$. How can I show that the functions $x \mapsto u(\lvert x\lvert)$ and $f'$ are uniformly continuous on $\mathbb{R}^d$?

My work so far:

The functions $u(\lvert x\lvert)$ and $f(x)$ are continuous on the compact set $C:=\{x\in\mathbb{R}^d:\lvert x\lvert\leq n\}$, hence uniformly continuous on that set. Let $\epsilon>0$ be given. There exist $\delta_1,\delta_2>0$ such that $x,y\in C$ and $|x-y|< \delta:= \text{min}(\delta_1,\delta_2)$ implies

$$|f(x)-f(y)|<\epsilon/2$$

$$|u(| x|)-u(| y|)|<\epsilon/(2||f||)$$

where $||f||:=\sup\{|f(x)|:x\in \mathbb{R}\}$. Now let $x,y\in \mathbb{R}^d$ with $|x-y|< \delta$. If $x,y\in C$ then

$$|f'(x)-f'(y)|\leq |u(| x|)||f(x)-f(y)|+|f(y)||u(| x|)-u(| y|)|<1\cdot \epsilon/2+||f||\cdot \epsilon/(2||f||)=\epsilon.$$

If $x\in C$ but $y\notin C$, then $|x|\leq n$ but $|y|>n$. Then the intermediate value theorem implies that there exist $t\in[0,1]$ such that $z:=ty+(1-t)x$ has norm $|z|=n$. Hence $z\in C$, $f'(z)=f'(y)=0$ and $|x-z|=|t||x-y|\leq|x-y|< \delta$, so the previous case implies $$|f'(x)-f'(y)|< \epsilon.$$

Finally if both $x,y\notin C$ then trivially $$|f'(x)-f'(y)|=0.$$

Is this correct?

Answer 1

It is much simpler:

If we set $\hat f:=u\cdot f$ then $\hat f\in C_b({\mathbb R})$, and $\hat f$ vanishes identically on ${\mathbb R}\setminus I_{n+1}$.

Claim: The function $\hat f$ is uniformly contiuous on ${\mathbb R}$.

Proof. Let an $\epsilon>0$ be given. As $\hat f$ is continuous on the compact set $I_{n+2}$ there is a $\delta\in\>]0,1[\>$, such that $|\hat f(x)-\hat f(y)|<\epsilon$ whenever $x$, $y\in I_{n+2}$ and $|x-y|<\delta$. In reality this $\delta$ works for all of ${\mathbb R}$: When $x, \>y\in{\mathbb R}$ have distance $<\delta<1$, and they are not both in $I_{n+2}$, then both are $\notin I_{n+1}$, hence $\hat f(x)=\hat f(y)=0$.

For the $d$-dimensional case replace $I_n$ by the $d$-dimensional ball $B_n:=\bigl\{x\in{\mathbb R}^d\bigm| \|x\|\leq n\bigr\}$.