What if any are the benefits, of presuming or deriving that a function:F
:F:[0,1] → [0,1] .
F being a tandard real valued function of one argument, -not a partial function, that assigns exactly one element of its co-domain, to each elements of its domain.
is 'strictly monotone increasing on its domain in contrast to 'monotone increasing (B): (A)∀ (x ∈ [0,1]=Dom(F): x>y ⟺ F(x)>F(y)
(B) (x ∈ [0,1]=Dom(F) x>y → F(x)≥F(y);
That is, as a tool in deriving that the function F is continuous or linear?
Particularly in the case, where a function is already quite homogeneous already?
For example, where∀ (x)∈ { Dℚ]∩ [0,1]}:F(x)=x.
where, F(1)=1 and F(0)=0, and Dℚ, denotes the set of all dyadic rational numbers∈[0,1]=Dom(F) ?
Tat is to show that∀ (x ∈ [0,1]=Dom (F) ): F(x)=x,?
Generally F need only be presumed, monotone increasing over its domain(F)=[0,1]. Where G is identity function, ∀ (x ∈ [0,1]∩ Dℚ]: F(x)=G(x)=x
That is, when dense set, arguments are applied to show that F=G(x)=x for all reals in [0,1] or (0,1)as, F=G, over all dyadic rationals, a dense subset ofdom(F)= [0,1], , and as G is is strictly monotone increasing and continous and as F is monotone increasing ,and preserves limits: Then generally it is concluded, that
∀ (x) ∈ (0,1) :F(x)=G(x)=x (for all reals in the open interval) and is continuous on (0,1).
Are there any benefits for these kinds of results, in already having that F is strictly monotone increasing (that is beforehand, and not as a consequence of the fact that F=G, and G the identity function, is strictly monotone increasing). Ie, to achieve said result. or something like it.?
s a strictly monotone increasing function that has a continuous first derivative such that F'(x)>0 as the derivative is defined insofar that F has a continuous derivative and where F' is defined the derivative is positive and dense in an interval for strictly monotone F, but as F' is continuous what holds in a dense subset holds for the entire domain . I presume not (i presume that only F'(0)=0 cant be dense in an interval); for example x^2 over [0,1] appears to have continuous derivative F'(x)=2x a linear function which is continuous and x^2 is strictly increasing yet clearly at x=0 F'(x)=0;
or is there some property that ensure this
as a monotone increasing first derivative of a twice continuosuly differentiable and convex and continuous function, F whose first t derivative is is positive almost every where said first derivaitive function F' itself is monotonic increasing due to convexity, a function that in fact actually has a strictly positive first derivative;
then if that derivative function F' agrees with some other strictly monotonic and increasing function G, over a dense subset,over the same domain, then F' is also strictly positive as F'=G, as F' is monotone and continuous and thus what applies to its F' derivatve ( its derivaitive agrees with some other function G that is continious and strictly monotonic over a dense subset) also applies to F'?
with the deriviative function, of some otherstrictly increasing continuious function whose first derivative happens to be also continuous, mononotonic increasing and positive everywhere, that the first derivative function are the same and so the original function F has a strictly positive first derivative
andso if that derivative function agrees some other derivative function that has a strictly positive first and monotonic increasing a continuous first derivative over the same domain, I poresume that the derivative would have to literlaly agree over that that dense subset, not just with regards positiveity
dense in an interval , then the original fuinction has what appears to be only a positive first derivative a.e that its first derivative function is montonic increasing itself, and continuous; that if its first derivative function is positive and agrees with
tslef ie its second derivative is non negative (and conve)
t and is monotone increasing and continuous first derivative of a function that F'(x)>0 over a dense subset,
as strictly increasing
. I presume the fact that a function F agrees with the identity function and F is strictly monotone increasing in dyadic rationals or rationals in [0,1] ie F(x)=x for dyadic rationals or rationals in [0,1]
somehow entailsF is strict monotonicit increasing given F is merely monotone increasing; can only always find for m1>m2 where m1,m2 are not element of the dense set, 2 element of the dense x,z,\in rationals for example, where x>z so that m1> x>z>m2,,
where as, x y in the dense set, F(x)=x, F(z)=z, so x> z implies F(x)=x>z=F(z)=z
F(x)>F(z)
so m2= F(x); implies F(m1)>=F(x)>F(z)>=F(m2)
so F(m1)>F(m2) in any case, ie F is strictly monotone.
Ie if F(x) is a monotonic increasing function and F(x) is strictly increasing over a dense subset of its domain, such as all of the rationals in [0,1] or dyadic rationals in [0,1] where
where [0,1] =dom(F), does this imply F injective and monotone and thus F strictly monotone increasing ?
can only guess that perhaps it has no real benefit here (when dealing with functions of one variable) unless one has not already have F(1)=1 and F(0)=0; apparently one there are some advantages about not having to worry about intricacies about sup, and inf F: at the end -piints, with strict monotone increasing-ness, but I presume that is is only an issue if one is attempting to derive those points already (as opposed to continuity at those points)?
It appears that most of the mathematics used to ensure that a function is the identity function and continuous as a side consequence, rely mainly on monotone increasing-ness-ness.
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