If $f$ is continuous then $f$ is uniformly continuous iff $|f|$ is uniformly continuous If $f:\Bbb R^n \to \Bbb R$ is continuous then $f$ is uniformly continuous iff $|f|$  is uniformly continuous.
A map $f$ from a metric space $M=(M,d)$ to a metric space $N=(N,\rho)$ is said to be uniformly continuous if for every $\epsilon>0$, there exists a $\delta>0$ such that $\rho(f(x),f(y))<\epsilon$ whenever $x,y \in M$ satisfy $d(x,y)<\delta$.
Clearly if $f:\Bbb R^n \to \Bbb R$ is uniformly continuous then $|f|$  is uniformly continuous as $|f|(x)-|f|(y)|\leq |f(x)-f(y)|$ but I am having a real trouble showing the converse part. In the region where $f$ is always positive or negative, we will not have any problem but how to deal with the points where $f$ is changing sign. If the zeros of $f$ are finite then also we can take a minimum of all $\delta$s and conclude the result. What will happen if zeros of $f$ are infinite?
 A: As mentioned in the comments, the proof given here may easily be modified to work for the whole of $\mathbb{R}^n$.
Since $\lvert f \rvert$ is uniformly continuous, there exists a $\delta > 0$ such that
\begin{align*}
d(x,y) \leq \delta \Rightarrow \lvert \lvert f \rvert (x) - \lvert f \rvert (y) \rvert \leq \frac{\epsilon}{2}.
\end{align*}
Note that if $f(x)f(y) > 0$, then
\begin{align*}
\lvert f(x)-f(y)\rvert
\leq \lvert \lvert f \rvert(x) - \lvert f \rvert (y) \rvert,
\end{align*}
which is less than $\epsilon/2$ whenever $d(x,y) \leq \delta$. Unsurprisingly, this case was quite trivial. We now turn our attention to the case where $f(x)f(y) \overset{\star}{\leq} 0$. Since it always holds that
\begin{align*}
\lvert f(x)-f(y)\rvert
\leq \lvert f \rvert(x) + \lvert f \rvert (y).
\end{align*}
it suffices to show that $\star$ implies the existence of a $z$ such that $d(x,z) \vee d(y,z) \leq d(x,y)$ and $f(z) = 0$. Because then
\begin{align*}
\lvert f(x) - f(y) \rvert
&\leq
\lvert \lvert f \rvert(x) - \lvert f \rvert(z) \rvert
+
\lvert \lvert f \rvert(y) - \lvert f \rvert(z) \rvert \\
&\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}
whenever $d(x,y) \leq \delta$. Since $f$ is continuous, the existence of a suitable $z$ follows from the continuity of $f$ and $\star$ (as a consequence of the Intermediate Value Theorem, see e.g. here).
