If $f_i$ has locally bounded variation for all $i$ and $(f_i)_i$ is summable, are we able to show that $\sum_{i∈I}f_i$ has locally bounded variation?

Question

Let

• $E$ be a normed $\mathbb R$-vector space
• $I$ be a nonempty set
• $f_i:[0,\infty)$ be of locally bounded variation$^1$ for $i\in I$

If $(f_i(t))_{i\in I}$ is summable$^2$ for all $i\in I$, are we able to show that $$f:=\sum_{i\in I}f_i$$ is of locally bounded variation?

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$^1$ Let $$\mathcal P_{[a,\:b]}:=\left\{(t_0,\ldots,t_k)\in\mathbb R^{k+1}:k\in\mathbb N\text{ and }a=t_0<\cdots<t_k=b\right\}$$ and $$|\varsigma|:=\max_{1\le i\le k}(t_i-t_{i-1})\;\;\;\text{for }\varsigma=(t_0,\ldots,t_k)\in\mathcal P_{[a,\:b]}$$ for $a,b\in\mathbb R$ with $a<b$. Let $$V_\varsigma(f):=\sum_{i=1}^k\left\|f(t_i)-f(t_{i-1})\right\|_E\;\;\;\text{for }\varsigma=(t_0,\ldots,t_k)\in\mathcal P_{[a,\:b]}$$ and $$V_{[a,\:b]}(f):=\sup_{\varsigma\:\in\:\mathcal P_{[a,\:b]}}V_\varsigma(f)$$ for $a,b\in\mathbb R$ with $a<b$ and $f:D\to E$ with $[a,b]\subseteq D\subseteq\mathbb R$. Let $a,b\in\mathbb R$ with $a<b$ and $f:D\to E$ with $[a,b]\subseteq D\subseteq\mathbb R$ $\Rightarrow$ $f$ is said to be of bounded variation on $[a,b]$ $:\Leftrightarrow$ $$V_{[a,\:b]}(f)<\infty\;.$$ $f:[0,\infty)\to E$ is said to be of locally bounded variation $:\Leftrightarrow$ $f$ is on bounded variation on $[0,t]$ for all $t>0$.

$^2$ $(x_i)_{i\in I}\subseteq E$ is called summable $:\Leftrightarrow$ $\exists x\in E$ with $$\forall\varepsilon>0:\exists J\subseteq I:|J|\in\mathbb N\Rightarrow\forall K\subseteq I:|K|\in\mathbb N\text{ and }J\subseteq K\Rightarrow\left\|x-\sum_{k\in K}x_k\right\|_E<\varepsilon\;.$$ In that case, $$\sum_{i\in I}x_i:=x\;.$$

Answer 1

Passing to a pointwise limit does not preserve being locally BV. Neither does a uniform limit, for that matter. An example would be $$ f_n(x) = \begin{cases} 0,\quad & x\le 2^{-n}, \\ x\sin(1/x),\quad & x>2^{-n} \end{cases} $$ These functions are locally BV for every $n$. They converge uniformly to $$ f(x) = x\sin(1/x) $$ which is not BV on $[0,1]$ (consider the partitions corresponding to maxima and minima of $\sin(1/x)$).

This can be restated in terms of infinite series $\sum (f_{n+1}-f_n)$ where the series converges absolutely in the uniform norm, $\|f_n-f_n\|\le 2^{-n}$; so it converges pointwise in whatever sense (absolutely, unconditionally).