a-continuous-function-which-is-not-of-bounded-variation

A continuous function which is not of bounded variation

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)=supPPnP1i=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

x is not an extreme point of CIt 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)=1u1 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

x is not an extreme point of CCan 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 p1, we have V10(g)>V2512p+1/2(g)=4(p1) and therefore V10(g)=+.

A continuous function which is not of bounded variation

x is not an extreme point of CAnd 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 x0 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 2p integer, we have V10(h)V2512p+1/2(h) as [12p+1/2,25][0,1]. Taking {12p+1/2,12p1/2,12p3/2,,12+1/2=25} as a partition of [12p+1/2,25], one can prove that V2512p+1/2(h)2p1i=11i+2. As the last series diverges, h is not of bounded variation.

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