A study on symmetry of second derivatives with symmetric functions
Abstract
The link between the two results is their equivalence, which is established in Theorem. For the sake of simplicity we confine our discussion to functions defined on open sets of the plane and to second order mixed partial derivatives. The interested reader can easily extend the obtained result to more general situations. All functions of two variables are assumed to be continuous (
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Introduction
Suppose is a real-valued function of two variables x,y and f(x,y)is defined on an open subset of U R 2 Suppose further that both the second-order mixed partial derivatives and exist and are continuous on . Then, we have: on all of .
General statement
The statement can be generalized in two ways:
We can generalize it to higher-order partial derivatives.
We can generalize it to functions of more than two variables.
Suppose is a function of two variables which we denote x and y. There are two possible second-order mixed partial derivative functions for , namely and . In most ordinary situations, these are equal by Clairaut's theorem on equality of mixed partials. Technically, however, they are defined somewhat differently. Often the term mixed partial is used as shorthand for the second-order mixed partial derivative. However, mixed partial may also refer more generally to a higher partial derivative that involves differentiation with respect to multiple variables.
Baire classification of symmetric derivatives
Let f be an extended real function, that is, a mapping from R into the extended real line. If there exists a sequence of continuous real functions, f 1 , f 2 , such that f n (x) →f(x) for each x R, then f is said to be in the first Baire class, which is denoted by B 1 . Theorem 2.1. If f , then f s B 1 Proof. Let If f and P be a perfect subset of R. Choose real numbers ,, and such that <<<. According to Preiss [14, Theorem 1], it suffices to show that the sets A = (x P: f s (x) ≤} and D = {x G P: f s (x) ≥} cannot both be dense in P. To do this, assume both A and D are dense in P in order to arrive at a contradiction. Define B = . Since P = B U C, at least one of the sets B or C must be a second category subset of P. Suppose first that B is a second category subset of P. Through the addition of an appropriate linear function to f, it may be assumed, without loss of generality, that ß = 0. Then define
Since B
B n and B is a second category subset of P, there exists a closed interval I and an integer n0 such that I n P is perfect and B n0 is dense in I P. Suppose . Then, since implies that there is a (x) (0, l/n 0 ) such that whenever Since I P is perfect, x is either a right or a left limit point of I P. We assume, without loss of generality, that x is a right limit point. Since A is dense in / n P, there exists a y A (x, x + (x)) and a (y) 0 such that when is also dense in I P, so a z may be chosen. Then, using the inequalities 0 <yz < (x), 0 < zx < (y) and 0 <yx < 1//n 0 we obtain the relations This obvious contradiction leads us to conclude that B cannot be a second category subset of P. A similar contradiction is reached if we assume that C is a second category subset of P. Thus, we are forced to conclude that A and D cannot both be dense in P. A more general derivative than the symmetric derivative could be defined as follows: Let a be a positive function defined on a neighbourhood of 0. Set and let be the class of all real functions for which exists everywhere. Then an argument similar to the proof of Theorem 2.1 shows that if/ , then/as . Although it is unknown whether each f is measurable, the next theorem does show something about the behavior of such a function and at the same time shows that not every function is a symmetric derivative
The structure of functions in m:
In this section we use the concepts of reflection and local symmetry. Suppose AR and x R. The reflection of A through x is given by . Then A is said to be locally symmetric at x if for some > 0 , or equivalently, if the characteristic function of A is symmetrically continuous at x. We refer to A as a locally symmetric set if it is locally symmetric at each x R. Lemma 4.1. f m, then the sets are both countable and closed.
Proof. We prove the lemma in the case of A. The assertion for B then follows by considering -f. It is clear that A is closed. Because C(f) is dense (Theorem 3.1), A must be nowhere dense. Since A is closed we may write A = P N, where P is perfect and N is countable. Suppose P , and let (, ) be a component of Pc ( or could be infinite). Since P , or must be finite. Suppose is finite. Then, since P is closed, P. P being perfect and , we see that for each > 0, is uncountable.
Since is countable, we may choose a sequence, , such that n decreases to . Because , we may choose a such that and . Consequently f . Similarly, if a , then .
Now we note that where each is a component of Pc. Since P is a nowhere dense set in R, the sets { n } and { n } are both dense in P. This implies, from the above, that the sets are disjoint and dense in P; furthermore, according to Theorem 2.1, I + and I -are G -sets. This contradicts the fact that P is a Baire space.
For any fm, we let We define f to be the real function We also define the extended symmetric derivative of f at x to be where (h 0) means that h 0 through the set . (Note that the extended symmetric derivative equals the symmetric derivative whenever the latter exists)
Conclusion
Jansen gave explicit formulas for the computation of the second-order derivatives of the elementary symmetric functions. But they are only applicable to those pairs of items which have unequal parameters. It is shown here that for equal parameters similar explicit formulas do exist, too, facilitating the application of the Newton-Raphson procedure to estimate the parameters in the Rasch model and related models according to the conditional maximum likelihood principle. A similar calculation for the other mixed partial yields that:
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