Property of periodic continuous mapping Let $f\ne0\in\mathbb{R}^\mathbb{R}$ be a continuous mapping with period $p>0$. Is it true that $id\cdot f$ is not uniformly continuous?
Let $a\in\mathbb{R}$ where $f(a)\ne0$ and $\delta>0$. If $1\leq p<\delta$, then $|a+p-a|<\delta$ and $|(a+p)f(a+p)-af(a)|=|pf(a)|\geq|f(a)|$. What to do $\delta\leq p$ or $p<1$? Any hints?

Assume that $f$ is not constant.  For any $\delta >0,$ fix $x,y$ with $|x-y|<\delta$ and $f(x)\ne f(y)$... (Note $x,y$ exist because $f$ is continuous but not constant)...
Now for any $n\in\Bbb N$ we have $|(x+np)-(y+np)|<\delta.$ But $$|(x+np)f(x+np)-(y+np)f(y+np)|=$$ $$=|np(f(x)-f(y))+(xf(x)-yf(y)|\ge$$ $$\ge np|f(x)-f(y)|-|xf(x)-yf(y)|,$$ which has no upper bound over the set of all $n\in\Bbb N.$
