MoaSticky NoteIf a curve is rectifiable, we can find the length of the curve by constructing an incribed polynomial in the curve and increase its number of edges to infinity. The sum of the lengths of the edges converges to the length of the curve. Evidently, each curve has a particular length (a positive real value) and the mapping of the curve to its length is a functional.
MoaSticky NoteThe definite integral alongside generates a particular real number depending on y(x). Thus, the integral maps a function to a contant and is a functional. All definite integrals in fact produce functionals so long as they exist.
MoaSticky NoteAssume a single function F. Notice how the definite integral value alonside is likely to fluctuate depending on the input y(x) and thus we have a functional. Note also how the functional in example 4 is a special case of this functional.
MoaSticky NoteThe Riemann integral is a sum. It is this fact which gives rise to the localization property mentioned alonside.
MoaSticky NoteNotice that in this case, this setup is almost a resetup of the n + 1th Riemann sum with an equally spaced partition with the exception that the derivative of the function is estimated by the average gradient. Evidently, the positioning of the polynomial lines depend on the function y(x). If we instead consider each output of y(x) to be a variable (as the function y(x) itself is considered to be variable), then this sum depends on the variables y1 to yn and we may thus optimize over those n variables.
We will class functions according to their properties.Furthermore, we will consider particular classes of functions depending on the problem at hand (e.g. a set of all continuous functions, a set of all differentiable functions, etc.)
MoaSticky NoteNotice that the objects in a linear space are not necessarily commutative nor is the concept of multiplication (and thus the concept of a multiplicative inverse) necessarily defined. Thus, a set of matrices for instance form a linear space.
MoaSticky NoteThe norm could be any function which satisfies these properties. It need not necessarily be the absolute value function. However, for a normed linear space to exist, some function must exist which satisfies these properties (and of course the objects must be partially ordered).
MoaSticky NoteIn every linear space, the definition of the norm (as well as addition and comparison) may differ. According to the definition of a norm in C, if the norm is a real number, the function must be bounded.
Because we will later define closeness according to the norm of a function, closeness in C implies that two functions both lie within the same band over some interval regardless of how small we make the band.
The contrapositive of this statement is easy to verify. If the functions or their derivatives exceed the band, the norm of the difference of the functions will also exceed the band.
MoaSticky NoteIt should be clear that if a function has an image in the real or complex number systems, all the axioms for a linear space will apply. If a function has a real image, the definition of a normed space will also obviously apply.
MoaSticky NoteThe continuity of a functional applies universally and is not restricted to normed linear spaces only.
In the defintion given alonside, y is a variable respresentative function for any function in the neighbourhood of y hat from the space which J operates over (this is indicated by delta). This is analogous to a variable being in some neighbourhood of the point of evaluation of a limit for functions (although the exact point need not lie in the domain of the function).
However, functions are not well ordered (they do not have a consecutive element). As such, continuity in this case is defined in such a way that the order of consideration of each function does not matter. We simply consider J to be continuous at some point in the function space if the image of J (for any representative function) is close to the image of J of a particular function when two functions are close (assessed by the norm of the space).
In terms of our definition of continuity of a functional, we need to ensure that J always exists. In the example given alonside, the function F may not exist as we cannot guarantee the existence of y' if y comes from C.
Thus, if we ensure that a functional is continuous, we automatically avoid this problem.
MoaSticky NoteA linear functional is a functional whose elements are from a normed linear space and satisfy the characteristics given alongside.
MoaSticky NoteAlthough continuous functions h(x) do exist such that alpha will be non-zero, the lemma stipulates that h(x) can be any continuous function and as such, the only way that we can guarantee that the integral sums to zero for any continuous function h is for alpha to be a zero function.
MoaSticky NoteNote the initial conditions h(a) = h(b) = 0. Let us interpret the derivative as the increment required at each point to bring about the function value at the next point divided by a tiny value. If we considered each tiny value to be equal (by creating an evenly spaced partition in [a, b], it should be evident that for the function to return to nil as given by the initial conditions, the increments must all sum to zero. If we factor the constant large factors out of the sum (integral), we will witness our result.
The lemma indicates that the function alpha must not interfere with each of the aforementioned increments.
MoaSticky NoteNotice that we could integrate the alonside expression by parts by deriving alpha and integrating h''(x) (if we impose the assumption that alpha is from space D1) in which case lemmas 1 and 2 come into play and the result will follow.
MoaSticky NoteWe can separate this integral as a sum of two integrals (this functional is in fact from the normed linear space) and then take one integral to the RHS. At this point, if we impose the restriction that alpha comes from D1, we could integrate by parts and obtain the result after some subtraction and inference. If we knew the result before hand, we could have reduced the intergrand to a derivative expression by undoing the product rule. If we then consider the integrand to be a product with 1, we would find lemma 2 at play.
A standard function is said to be differentiable if we can find its differentiable (although the word is used in other contexts as well).
We consider a functional to be differentiable if the functional can be rewritten in terms of a linear functional plus a tiny constant which gets smaller as the norm of the function increment tends to zero (it makes sense to add a constant as part of the representation as the functional generates real values). Note that epsilon itself gets smaller as ||h|| tends to 0. If we interpret the norm as the size of increment, this means that we do not cause the original function to deviate too much from its initial point. Thus the resulting difference from the starting point would also tend toward 0.
The linear functional portion of the representation is called the differential of the functional. It is dependent upon a preselected value of y and h.
Note that constant part of the representation depends on the product of both ||h|| and epsilon and as such is considered to be of degree greater than 1 relative to ||h||.
Note that by definition, the differential is a linear functional. This guarantees us that h is from C. However, h could be a vast number of functions. Thus the only way that we can guarantee that the differential tends to 0 for all h is if the differential as a function of h is always 0. Thus the differential functional must be selected so that when it takes in h as a parameter, its value is always 0.