Let a real-valued function $f(x)$ defined on $\mathbb{R}$.
- For a bounded interval $[a,b]\subset \mathbb{R},$ taking a partition $\mathcal{P} =\{x_0, x_1, \ldots, x_n\},$ where $a\le x_0<x_1<\cdots<x_n\le b,$ then we have corresponding $V^{\mathcal{P}}(f):=\sum_{i=1}^{n} |f(x_i) - f(x_{i-1})|.$ If the set $\{V^\mathcal{P}(f) : \mathcal{P} \text{ is a partition of } [a, b]\}$ is bounded, then $f$ is called a function of bounded variation on $[a, b]$ and $V_{[a, b]}(f) := \sup_\mathcal{P} V^\mathcal{P}(f)$ is the total variation of $f$ on $[a, b]$.
Similarly,
- For the entire $\mathbb{R}$,if the set $\{V_{[a,b]}(f) : [a, b] \text{ is a bounded interval on } \mathbb{R} \}$ is bounded, then $f$ is called a function of bounded variation on $\mathbb{R}$ and $V_{\mathbb{R}}(f) := \sup_{[a,b]} V_{[a,b]}(f)$ is the total variation of $f$ on $\mathbb{R}$.
Consider $f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}},\sigma>0,x\in(-\infty,\infty).$
For any bounded interval $[a,b]$,we can get that $f:[a,b]\rightarrow \mathbb{R}$ is Lipschitz continuous on $[a,b]. \Rightarrow f$ is absolutely continuous on $[a,b].\Rightarrow f $ is of bounded variation on $[a,b].$
But I am not sure $f$ is of bounded variation on $\mathbb{R}.$ Intuitively, this doesn't seem to hold,but how to rigorously prove it is unbounded variation on $\mathbb{R}？$