Volterra's function: Difference between revisions

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==Definition and construction==

The function is defined by making use of the [[Smith–Volterra–Cantor set]] and "copies" of the function defined by ''f''(''x'') = ''x''²sin(1/''x'') for ''x'' ≠ 0 and ''f''(''x'') = 0 for ''x'' = 0. The construction of ''V''(''x'') begins by determining the largest value of ''x'' in the interval [0, 1/8] for which ''f'' ′(''x'') = 0. Once this value (say ''x''₀) is determined, extend the function to the right with a constant value of ''f''(''x''₀) up to and including the point 1/8. Once this is done, a mirror image of the function can be created starting at the point 1/4 and extending downward towards 0. This function, which we call ''f''₁(''x''), will be defined to be 0 outside of the interval [0, 1/4]. We then translate this function to the interval [3/8, 5/8] so that the function is nonzero only on the middle interval as removed by the [[Smith-Volterra-Cantor setSmith–Volterra–Cantor|SVC]]. To construct ''f''₂(''x''), ''f'' ′(''x'') is then considered on the smaller interval 1/16 and two translated copies of the resulting function are added to ''f''₁(''x''). Volterra's function then results by repeating this procedure for every interval removed in the construction of the [[Smith-Volterra-CantorSmith–Volterra–Cantor set|SVC]].

==Further properties==

Volterra's function is differentiable everywhere just as ''f''(''x'') (defined above) is. The derivative ''V'' &prime;(''x'') is discontinuous at the endpoints of every interval removed in the construction of the [[Smith-Volterra-CantorSmith–Volterra–Cantor set|SVC]], but the function is differentiable at these points with value 0. Furthermore, in any neighbourhood of such a point there are points where ''V'' &prime;(''x'') takes values 1 and &minus;1. It follows that it is not possible, for every ε > 0, to find a partition of the real line such that |''V'' &prime;(''x''₂) &minus; ''V'' &prime;(''x''₁)| < ε on every interval [''x''₁, ''x''₂] of the partition. Therefore, the derivative ''V'' &prime;(''x'') is not Riemann integrable.

A real-valued function is Riemann integrable if and only if it is bounded and continuous almost-everywhere (''i.e.'' everywhere except a set of [[measure theory|measure]] 0). Since ''V'' &prime;(''x'') is bounded, it follows that it must be discontinuous on a set of positive measure, so in particular the derivative of ''V''(''x'') is discontinuous at uncountably many points.