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I have this problem, Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges at each point of $[a,b]$ to a function $f$. Then $T_a^b(f)≤\liminf T_a^b(f_n)$.

It is solved in, Pointwise convergent and total variation.

But I have a question about the answer of Nana (the last one) she states:

Hint:

Let $a=x_0\lt \ldots \lt x_n=b$ be a subdivision of $[a,b]$. Let $ε>0$. Then there is an $N$ such that $$|f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2$$

Then consider

$$\sum_{k=1}^n |f(x_k) - f(x_{k-1})|\leq \sum_{k=1}^n |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^n |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})|$$

Can you continue?

My question is about how to choose of the epsilon, because if $$|f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2$$

then $$\sum_{k=1}^n |f(x_k) - f(x_{k-1})|\leq \sum_{k=1}^n \varepsilon +\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})| \\ = n\varepsilon +\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})| \leq n\varepsilon +T_a^b(fn)$$

and the what? I know we want to get $T_a^b(f)\leq \varepsilon +T_a^b(fn)$, but I don't understand how, please help...

Community
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Rafael
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2 Answers2

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The problem is that the index $n$ of the function $f_n$ shouldn't be the same as that of the number of elements of the partition. We assume that the partition is $a=x_0<\dots <x_K=b$, and the inequality should read $$\sum_{k=1}^K |f(x_k) - f(x_{k-1})|\leq \sum_{k=1}^K |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^K |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^K |f_n(x_k) - f_n(x_{k-1})|.$$

Davide Giraudo
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  • ok that help, but the main problem still the same, because then $$\sum_{k=1}^K|f(x_k)-f(x_{k-1})|\leq K\varepsilon+\sum_{k=1}^K|fn(x_k)-fn(x_{k-1})|$$ and when we take supremun to get the total variation, $K\rightarrow \infty$. would that means that $T_a^b(f)$ is unbounded? or not? – Rafael Nov 22 '14 at 20:27
  • Once you get your displayed inequality, you can bound the second term of the RHS by the total variation of $f_n$. The therm $K\varepsilon$ can be taken as $\varepsilon$ instead. – Davide Giraudo Nov 22 '14 at 20:55
  • There is a similar problem in Folland (problem 3.32) but the functions are defined on all of $\mathbb{R}$ and they are assumed to live in NBV (normalized bounded variation). I have not been able to figure out why this last condition is required as. Does anyone have any ideas? – unknownymous Nov 27 '14 at 05:01
  • @unknownymous The reason for the condition that they are assume to be NBV is that this way they are all linear functionals on the continuous function on $\mathbb{R}$. So that we can use Lax Functional Analysis chapter 10 theorem 5 to prove the inequality. – Andrew Liu Sep 27 '16 at 18:43
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Let $f_n$ be a sequence of functions on $[a,b]$ that converges at each point of $[a,b]$ to $f$. Let $a=x_0<x_1<...x_n=b$ be a partition of $[a,b]$ and let $\epsilon > 0$. Then there exists $N$ such that $$\sum\limits_{i=1}^n |f(x_i) -f(x_{i-1})| \leq \sum\limits_{i=1}^n |f(x_i) - f_n(x_i)| +\sum\limits_{i=1}^n |f(x_{i-1}) - f_n(x_{i-1})| +\sum\limits_{i=1}^n |f_n(x_i) -f_n(x_{i-1})| < \epsilon +T_a^b(f_n)$$ for $n \geq N$. Thus $$T_a^b(f) \leq \epsilon +T_a^b(f) \ for \ n \geq N$$ so $$T_a^b(f) \leq \epsilon + \underline{\lim} T_a^b(f_n)$$ Since $\epsilon $ is arbitrary, $$T_a^b(f) \leq \underline{\lim} T_a^b(f_n)$$

Lucas
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