# Total variation of f on [0,1]

1. Aug 17, 2007

### bham10246

1. The problem statement, all variables and given/known data
Let $f_n, f: [0,1]\rightarrow \mathbb{R}$, and $f_n(x)\rightarrow f(x)$ for each $x \in [0,1]$.

I need to show the following two things:
a. $T_0^1(f)\leq \lim\inf_{n\rightarrow \infty} T_0^1(f_n)$, and

b. if each $f_n$ is absolutely continuous and $T_0^1(f_n)\leq 1$ for each $n$, then $T_0^1(f) = \lim_{n\rightarrow \infty} T_0^1(f_n)$.

2. Relevant equations

We denote $T_0^1(f)$ as the total variation of f on $[0,1]$.

3. The attempt at a solution

a.
If $f_n$ is not of bounded variation, then $T_0^1(f_n)=\infty$ and we're done.

So assume $f_n$ is of bounded variation. Then since $f_n(x) \rightarrow f(x)$ for each x, then for a partition $0=t_0 < t_1 < ... < t_N= 1$,

$|f_n(t_j)-f_n(t_{j-1})| \rightarrow |f(t_j)-f(t_{j-1})|$.

So $\sum_j|f_n(t_j)-f_n(t_{j-1})| \rightarrow \sum_j |f(t_j)-f(t_{j-1})|$.
Take the sup from both sides and so we have $T_0^1(f_n)\rightarrow T_0^1(f)$.

I would like to use Fatou's at some point but I would like some hints on how to change the idea of total variation into a sequence of nonnegative measurable functions....

2. Aug 17, 2007

### EnumaElish

fn are bounded. Let L be a lower bound for all n.

If L > 0, then all fn are positive.

If L < 0, then you can define gn = fn - L > 0.

If fn is (abs.) cont. then it is measurable (i.e. Borel).

Last edited: Aug 17, 2007
3. Aug 17, 2007

### bham10246

Definition of Total Variation: Let f(t) be complex-valued function defined on the interval [0,1]. Let $P: 0=t_0 < t_1 < ... < t_N = 1$ be a partition of the unit interval. Then
$T_0^1(f) = \sup_{P} \:\sum_{i} |f(t_i)-f(t_i-1)|$
where we take the supremum over all partitions of the unit interval.

Fatou's Lemma: Let $\{f_n\}_n$ be a sequence of nonnegative measurable functions on [0,1]. Assume $f_n(x)\rightarrow f(x)$ for almost all x. Then
$\int f \leq \lim\inf_{n\rightarrow \infty} \int f_n$.

Last edited: Aug 17, 2007