Holder continuity

There are associated concepts of left-continuity and right-continuity. 

The value of the derivative exists at x provided that we can choose some finite value c such that for all Sx small enough 

Note the difference between this and the definition of continuity, above. It is quite possible for a continuous function not to have a derivative. f(x) = a 1/3 is a good example. 

The first differences on both left and right of x = 0 tend to zero, but the first divided differences on both sides diverge. 

The second derivative is just the derivative of the derivative, and the nth derivative is defined by recursion on n. 

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The first form of this question is Is the function continuous and if the answer is yes , it is rapidly followed up by How many continuous derivatives does it have . In fact for really interesting definitions it is possible to ask also, Just how continuous is the highest continuous derivative , and that is expressed in terms of the Holder continuity exponent. 

The Holder continuity is a pair of values, i, f. The first of these, i is integral, and is the number of derivatives which exist, the second, /, in the closed interval (0. .. 1) which is essentially the value such that the ith derivative behaves like xJ as x approaches zero from above. These two numbers are sometimes written as i + /, sometimes as i, f but both indicate the same level of continuity. This leads to the notation C i+f where the + is explicit. 

The above defines the Holder continuity at a point. The Holder continuity of a complete curve is the lowest pointwise continuity. 

Thus the analysis of this chapter has indeed determined the Holder continuity, but only of the basis function and only at its end points. There is much more to do before we have a full answer to the continuity question. 

The earliest work on continuity focussed on the questions here is a scheme is it continuous is it C1 is it C2 which can be subsumed into the single question How many derivatives are continuous We now phrase the question a different way what are upper and lower bounds on the Holder continuity This is because numbers like — log( / )/log(a) are very rarely integers. 

The number found in the previous chapter is a strict upper bound on the Holder continuity because we have an example of a discontinuity at one particular place of one particular limit curve. 

Again we have constructed a discontinuity which can be used as an upper bound, in this case 2+1, on the Holder continuity. 

In general every power of the scheme considered introduces new mark points, and there is no guarantee in general that the 117th power will not show us places where the Holder continuity is lower than that found for lower powers. Thus this procedure can only ever give us upper bounds on the Holder continuity. 

In some rather special cases, it is possible for the matrices on some diagonals to have only polynomial eigencomponents, saying that the Holder continuity is infinite. This happens for the B-splines. 

However, if we are ever to say with confidence This is the Holder continuity of curves generated by this scheme , we need also lower bounds, which we can approach rigorously by using difference schemes. 

We can keep going, taking higher and higher divided difference schemes, until we find a scheme whose difference scheme is not contractive. The last scheme which was continuous gives a lower bound on the Holder continuity of the original scheme. 

However, the bounds determined in this way are only lower bounds. It is possible for a scheme to fail at a certain level, even when the Holder continuity is actually higher. Just as in the eigenanalysis case, we can often tighten the bounds by taking a power of the scheme. 

An upper bound on the Holder continuity of S is k — loga( e )/n and a lower bound k — log0( Z )/n, except that if either bound is an integer i the Holder bound is — 1,1 rather than i. 

The Holder continuity is a function of the scheme and therefore of the coefficients when a scheme is expressed as a linear combination of B-splines. For schemes with small kernels it is possible to use sharp specific arguments to determine this directly, at least for certain ranges of coefficients. We saw above that the continuity degree was determined by the kernel, almost independently of the number of a factors, which merely added a separate term. 

Instead of starting with low degree refinements, carry out the first hundred of whatever scheme you like to name - let s say the four-point scheme. Then after one hundred steps switch to the UP rules. The result will be your scheme for all practical purposes, but the limit curve has infinite continuity. This is an argument which says that Holder continuity of itself is not an important criterion. 

The first of these has exactly the same eigenvectors as the binary quadratic scheme of the previous question, but eigenvalues 1,1/3,1/9,1/9. Again, the fourth of the column eigenvectors is not polynomial, and the associated eigenvalue is 1/9 and so the Holder continuity is no better than —log3( 1/9) = 1 + 1. 

The column eigenvectors are all polynomial, and so there is no constraint on the Holder continuity here. 

The third column eigenvector is non-polynomial, and so the Holder continuity is no higher than —logz 1/6) ps 1 + 0.65. 

There is a Jordan block from the two 1/4 eigenvalues, which have a coupling value of 1. This causes a new component of the second difference (of size proportional to the fourth difference) to be added at each refinement. Thus the second difference grows arithmetically, and the Holder continuity is no better there than 1+1. 

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