Uniform continuity and the sup norm

[AxiomOfChoice]

Suppose I have a function f(x) \in C_0^\infty(\mathbb R), the real-valued, infinitely differentiable functions with compact support. Here are a few questions:

(1) The function f is trivially uniformly continuous on its support, but is it necessarily uniformly continuous on \mathbb R?
(2) I want
<br /> \left\| f(x-t) - f \left(\frac{x}{1+\epsilon} - t \right) \right\|_\infty \to 0<br />
as \epsilon \to 0. Is that so? It seems so, intuitively, if f is uniformly continuous.
(3) Also, is it true that
<br /> \left\| f(x-t) - f \left(\frac{x}{1+\epsilon} - t \right) \right\|_\infty = \left\| f(x) - f \left(\frac{x}{1+\epsilon} \right) \right\|_\infty<br />

 

[AxiomOfChoice]

I believe I have found the answer to (1) here, and it is yes, but the proof isn't exactly trivial.

 

[jbunniii]

Suppose I have a function f(x) \in C_0^\infty(\mathbb R), the real-valued, infinitely differentiable functions with compact support. Here are a few questions:

(1) The function f is trivially uniformly continuous on its support, but is it necessarily uniformly continuous on \mathbb R?

Yes. Suppose the support of $f$ is $K$. Since $K$ is compact, it is bounded, so it is contained within some closed interval $I$ with radius $r$, centered at the origin. Now expand this to a closed interval $J$ with radius $r+1$, centered at the origin. Since $f$ is continuous on the compact set $J$, it is uniformly continuous on that set. Let $\epsilon > 0$. There exists $\delta > 0$ such that whenever $x,y \in J$ with $|x-y| < \delta$, we have $|f(x)-f(y)| < \epsilon$.

Now let $\delta_0 = \min(\delta, 1)$, and consider any points $x,y \in \mathbb{R}$ with $|x-y| < \delta_0$. If $x,y\in J$ then $|f(x) - f(y)| < \epsilon$ because of the uniform continuity. If $x,y\in J^c$ then $f(x) = f(y) = 0$ so surely $|f(x)-f(y)| < \epsilon$.

The only other possibility is that one of the points (say $x$) is in $J$ and the other ($y$) is in $J^c$. Once again this gives us $f(x) = f(y) = 0$, because $|x-y| < 1$ and $y \in J^c$ imply that $x \in J \setminus I$, so $x$ is outside the support of $f$. Thus in this case we also have $|f(x) - f(y)| < \epsilon$.