Volterra's function
In mathematics, Volterra's function, named for Vito Volterra, is a real-valued function V(x) defined on the real line R with the following curious combination of properties:
- V(x) is differentiable everywhere
- The derivative V ′(x) is bounded everywhere
- The derivative is not Riemann-integrable.
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) = x2sin(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 x0) is determined, extend the function to the right with a constant value of f(x0) 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 f1(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 SVC. To construct f2(x), f ′(x) is then considered on the smaller interval 1/16 and two translated copies of the resulting function are added to f1(x). Volterra's function then results by repeating this procedure for every interval removed in the construction of the SVC.
Further properties
Volterra's function is differentiable everywhere just as f(x) (defined above) is. The derivative V ′(x) is discontinuous at the endpoints of every interval removed in the construction of the 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 ′(x) takes values 1 and −1. It follows that it is not possible, for every ε > 0, to find a partition of the real line such that |V ′(x2) − V ′(x1)| < ε on every interval [x1, x2] of the partition. Therefore, the derivative V ′(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 0). Since V ′(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.