Prove TV(f) ≤ lim inf TV(fn): Total Variation Homework

[sbashrawi]

Homework Statement

Let {fn} be a sequence of real-valued functions on [a, b] that
converges pointwise on [a, b] to the real-valued function f. Show that
TV (f) <= lim inf TV (fn).

Homework Equations

The Attempt at a Solution

I prove it in the follwoing way:
|fn(xi) - f(xi) | = 0 as n goes to infinity, then
| sum ( fn(xi) - f(xi)| goes to zero
| sum [ fn(xi) - fn(x_i-1)] - sun [ f(xi) - f(x_i-1)]| goes to zero
so the total variation TV(fn) converges to TV(f) point wise
hence TV(f) <= lim inf TV(fn)

but I am not sure about it since there is no condition that gurantees the concolusion

 

[mighty2000]

that the difference of TV is zero as n goes to infinity.

Thank you for your post. Your solution is on the right track, but there are a few details that need to be clarified.

Firstly, the statement "TV(fn) converges to TV(f) pointwise" is not entirely accurate. The total variation of a function is a real number, not a function, so it cannot converge pointwise. What you mean to say is that the sequence of total variations {TV(fn)} converges to TV(f), which is a real number.

Secondly, your argument about the difference of TV being zero as n goes to infinity is not completely justified. While it is true that the difference of TV(fn) and TV(f) will approach zero as n goes to infinity, this does not necessarily mean that the difference itself will be zero. In order to prove this, you need to use the fact that the sequence of functions {fn} converges pointwise to f. This means that for any given x in [a,b], the values of fn(x) will get closer and closer to f(x) as n goes to infinity. Therefore, the difference between fn(xi) and f(xi) will approach zero as n goes to infinity, and this will hold for all xi in [a,b].

To make this argument more rigorous, you can use the definition of pointwise convergence and the triangle inequality to show that for any given ε > 0, there exists an N ∈ ℕ such that for all n ≥ N and all x ∈ [a,b], |fn(x) - f(x)| < ε. From here, you can use the fact that TV is a non-negative function and apply the triangle inequality again to get |TV(fn) - TV(f)| ≤ TV(|fn - f|) < ε(b-a). This shows that the difference between TV(fn) and TV(f) can be made arbitrarily small, and therefore it must be equal to zero.

Once you have established that the difference between TV(fn) and TV(f) is zero, you can conclude that TV(f) ≤ lim inf TV(fn) as you did in your solution. This is because for any given ε > 0, there exists an N ∈ ℕ such that for all n ≥ N, TV(fn) < TV(f) + ε. Taking the limit as n goes to infinity, you get lim inf TV(fn) ≤ TV(f) + ε