Let $f : \Bbb R \longrightarrow \Bbb R$ be defined by $f(x) = \displaystyle {\sum\limits_{n = 1}^{\infty} f_n (x)},$ where $$ f_n(x) = \left\{ \begin{array}{ll} n \left (x - n + \dfrac 1 n \right ), & \quad x \in \left [n - \dfrac 1 n, n \right ], \\ n \left (n + \dfrac 1 n - x \right ), & \quad x \in \left [n, n + \dfrac 1 n \right ], \\ 0, & \quad \text {otherwise}. \end{array} \right. $$
Is $f$ uniformly continuous on $\Bbb R\ $?
What I observe is that if $n \geq 3$ then for all $x \in \left [n - \dfrac 1 n, n + \dfrac 1 n \right ]$ we have $f(x) = f_n (x).$ Now if $f$ is uniformly continuous on $\Bbb R$ then it is uniformly continuous on $\displaystyle \bigcup\limits_{n=3}^{\infty} \left (n - \dfrac 1 n, n \right ).$ Let us assume that $f$ is uniformly continuous on $\Bbb R.$ Now let us take $\varepsilon = \dfrac 1 5.$ Then there exists $\delta \gt 0$ such that for $x,y \in \displaystyle \bigcup\limits_{n=3}^{\infty} \left (n - \dfrac 1 n, n \right )$ with $|x - y| \lt \delta$ we have $|f(x) - f(y) | \lt \dfrac 1 5.$ But then for all $n \geq 3$ with $x,y \in \left (n - \dfrac 1 n, n \right )$ and $|x - y| \lt \delta$ we have $|f(x) - f(y)| \lt \dfrac 1 5.$ Take $N \in \Bbb N$ with $N \geq 3$ such that $\dfrac 1 N \lt \delta.$ Now for $x,y \in \left (N - \dfrac 1 N, N \right )$ with $|x - y | = \dfrac {1} {4N} \lt \dfrac 1 N \lt \delta\ $ (for instance we may take $x = N - \dfrac {1} {2N}$ and $y = N - \dfrac {1} {4N}$) we have $$|f(x) - f(y)| = N\ |x - y| = \dfrac 1 4 \not\lt \dfrac 1 5. $$ Hence $f$ can't be uniformly continuous on $\Bbb R.$
Can anybody please check my reasoning? Thanks in advance.