While preparing to test in calculus I found the question above:
Let $f_n(x)\to f$ uniformlly on $[a,b]$. Prove or give counterexample: if $\forall n\in\mathbb N f_n(x)\in BV$ and $\exists M>0$ s.t $\displaystyle V_a^b f_n\le M$ then $f\in BV$
I am really ashamed to say I don't have any ideas how to solve the question. Even though I think it's correct. First I think we need to define a partition on $[a,b]$ by $x_0<x_1<...<x_n$. Since $\forall n,f_n\in BV, |f_n(x_i)-f_n(x_{i-1})|<V_a^bf_n\le M$ Then we need to prove $V_{a}^bf=q<\infty$, which means taking sup for the expressions (created by paritions) in form of $\displaystyle\sum_{i=0}^{n}|f(x_i)-f(x_{i-1})|$. I thought changing the expression to $$\displaystyle\sum_{i=0}^{n}|f(x_i)-f_n(x_i)+f_n(x_i)-f(x_{i-1})|\le \sum_{i=0}^{n}|f_n(x_i)-f_(x_i)|+|f_n(x_i)-f(x_{i-1})|$$. The first part is final and defined but I don't know how to manage with the second part ($|f_n(x_i)-f(x_{i-1})|$). How can I solve the question?
Take limits instead. Let $x_0 < \cdots < x_k$ be a partition of $[a,b]$. Then $$ \sum_{i=1}^k |f_n(x_{i}) - f_n(x_{i-1})| \le V_a^b f_n \le M $$ for every index $n$, which implies that $$ \sum_{i=1}^k |f(x_{i}) - f(x_{i-1})| \le M $$ by taking the limit as $n \to \infty$ of the finite sum on the left. Note that $f_n \to f$ pointwise is all that is required. Now take the supremum over all partitions to get $V_a^b f \le M$.