I recently encountered a problem in one of my lectures about uniformly continuous functions, but I'm still relatively new to the concept and would to some feedback on my attempt
Problem statement: Let $f$ be a continuous function on an interval such that $|f|$ is uniformly continuous. Prove or disprove that $f$ is also uniformly continuous.
Attempted solution:
Let $\epsilon > 0$ be given. By the definition of uniform continuity, there exists a $\delta > 0$ such that for any $x,y$ in the interval with $|x-y| < \delta$, we have $||f(x)| - |f(y)|| < \epsilon/2$.
Then, consider the following cases:
If $f(x), f(y)$ are both positive, then $|f(x) - f(y)| = ||f(x)| - |f(y)|| < \epsilon/2 < \epsilon$.
If $f(x), f(y)$ are both negative, then $|f(x) - f(y)| = |-|f(x)| + |-f(y)|| = ||f(x)| - |f(y)|| < \epsilon/2 < \epsilon$.
If $f(x) < 0 < f(y)$, then by the continuity of $f$, there exists a point $c$ between $x$ and $y$ such that $f(c) = 0$. Moreover, since $|x-c| < \delta$ and $|y-c| < \delta$.
Therefore, $|f(x) - f(y)| = |f(x) - f(c) + f(c) - f(y)| \leq ||f(x)| - |f(c)|| + ||f(y)| - |f(x)|| < \epsilon$.
Hence, we have shown that for any $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < \delta$, which proves that $f$ is uniformly continuous.
Am i missing some thing here?
I think there might be a shortcut. If $f$ is uniformly continuous then $\forall k \in R, kf$ is uniformly continuous. If $f>0$ then $|f|=1\cdot f$. If $f<0$, then $|f|=(-1)\cdot f$. So $|f|>0 \implies |f|$ is uniformly continuous.
Suppose $f(c)=0$ and $f$ is not uniformly continuous at 0. Doesn't that imply that its not uniformly continuous near zero either? If so, then we have a contradiction, so $f$ is UC where $f(c)=0$ as well.