Introduction on total variation of functions
Recall that a function of bounded variation, also known as a BV-function, is a real-valued function whose total variation is bounded (finite).
Being more formal, the total variation of a real-valued function f, defined on an interval [a,b]⊂R is the quantity:
Vba(f)=supP∈PnP−1∑i=0|f(xi+1)–f(xi)| where the supremum is taken over the set P of all partitions of the interval considered.
First example of a function which is not of bounded variation
It is easy to show a function that is not of bounded variation. Consider the function f defined on the interval [0,1] by:
f(x)={0if x=01xif x∈(0,1]
For 0<u<1, we have V10(f)>V1u(f)=1u–1 and taking u as small as desired we get V10(f)=+∞. The function f is not bounded on its domain.
A bounded function which is not of bounded variation
Can a bounded function not be of bounded variation? The answer is also positive as proven by taking g defined on the interval [0,1] by:
g(x)={0if x=0sin(πx)if x∈(0,1]
For an integer p≥1, we have V10(g)>V2512p+1/2(g)=4(p−1) and therefore V10(g)=+∞.
A continuous function which is not of bounded variation
And finally we prove that the function h defined on the interval [0,1] by:
h(x)={0if x=0xsin(πx)if x∈(0,1]
is continuous and not of bounded variation. Indeed h is continuous at x≠0 as it is the product of two continuous functions at that point. h is also continuous at 0 because |h(x)|≤x for x∈[0,1]. We now consider the total variation of h. For 2≤p integer, we have V10(h)≥V2512p+1/2(h) as [12p+1/2,25]⊂[0,1]. Taking {12p+1/2,12p−1/2,12p−3/2,…,12+1/2=25} as a partition of [12p+1/2,25], one can prove that V2512p+1/2(h)≥2∑p−1i=11i+2. As the last series diverges, h is not of bounded variation.