- Can a Continuous, and (twice differentiable)Strictly (monotone)Increasing Function F.
F:[0,1] to [0,1]; F(1)=1 F(0)=0, F(0.5).
1.can F both strictly pseudo-concave and strictly pseudo convex.
2.Can F both be Convex, and strictly pseudo concave
3. F(x)=x which is concave and convex, strictly pseudo concave and strictly pseudo convex
Maybe you could try clarifying which part is your actual question. It's not clear what the hanging F(0.5) is doing in your question above. Also, since you did not specify, I'll just assume we are operated in R1, but this should work in any finite dimensional real space RN. Also, I'm ignoring the F(0.5) in my answer.
(1) is true for F(x)=x. F(x) is strictly pseudoconvex, because for any z not equal y, (y-z)*\grad{F(z)} >= 0 implies F(y) > F(z). Why? \grad{F(z)} is clearly just 1. So z not equal y and (y-z) >= 0 actually implies (z-x)>0, so z > x. So we need to show y>x => F(y)>F(x), which is obviously true if F(x)=x.
(2) F(x)=x is convex. F(x)=x is also strictly concave.
Nothing to add on Stefan's contribution since any linear function is both convex and concave. The being pseudo strictly requirement is simply fulfilled by the identity map.
Yes i though so. Any strictly increasing, pseudo linear function that is at least possesses a first derivative, I think is strictly both (strictly pseudo-linear)
Iget the impression that strict pseudo-linearity, (strict pseudo concave and strict pseid convexity) which is generally implied by pseudo convexity and pseudo concavity and strict quasi convexity and strict quasi concavity will be satisfied by any linear model generally, where the strict quasi convexity and strict quasi concavity are implied by strict monotincity-
So am I correct in presuming that the only advantage of adding pseo-convexity and pseudo concavity to a strictly increasing function is to ensure that its first derivative is always positive; ie if the derivative is ever zero then this must be local min and maxima, which strictly monotonic functions do not generally have (there are cases where strictly increasing functions have zero, first derivative in many places)
@Stefan I presume you mean that F(x) is convex and strictly pseudoconvex. Its not strictly convex? It is is stricly monotonic increasing and convex where strict monotone increasing generally implies strict quasi convexity and strict quasi concavity- and the linearity of the funtion implies convexity and concavity which imply pseudo-convex and pseudo concavity; so its strictly both. I presume that its generally just about any, a strictly monotonically increasing , continuous,and is differen-tiable; where this first derivative>0e. Is strictly both.
I think that the only advantage that a function has, in being a first /twice differen-tiable, continuous and pseudo linear, is that if it is strict monotone increasing functions, all it does, I presume, is out the degenerate cases where the first derivative is zero (these are just the cases of strictly quas-ilinear function, which have zero positive derivative at some x, , yet at x, F(x) is not at a local minima or maxima at that point (as strictly monotone increasing function do not usually have such pts in most normal cases )
Yes. Point (2) above was meant to be:
(2) F(x)=x is convex. F(x)=x is also strictly pseudoconcave.
yes its strictly pseudo convex and strictly pseudo concave- My main question is whether these constraints will ensure that a strictly monotonic increasing function on real interval to a real interval has a strictly positive derivative (if differentiable). My understanding is that strict monotone increasing, implies all of the standard but strictest quasi's convexity and quasi concavity requirements by definition?(strict, semi-strict standard quasi convexity, and strict, semi-strict and standard quasi concavity)
F(x)=x is concave and convex, strictly pseudo concave and strictly pseudo convex.
The following question arises: Characterize functions F:[0,1] to [0,1] which are
both strictly pseudo concave and strictly pseudo convex on I=[0,1] ?
Recall:
f is said to be pseudoconvex if, for all x, y in I=[0,1] such that f'(x)\cdot (y-x)\geq 0, we have f(y) \geq f (x).
Every convex function is pseudoconvex, but the converse is not true. For example, the function ƒ(x) = x + x^3 is pseudoconvex but not convex. Any pseudoconvex function is quasiconvex, but the converse is not true since the function ƒ(x) = x^3 is quasiconvex but not pseudoconvex. Pseudoconvexity is primarily of interest because a point x* is a local minimum of a pseudoconvex differentiable function ƒ if and only if it is a stationary point of ƒ.
Not all convex functions are strictly pseudo convex apparently or strictly quasi convex in the strongest sense. I think that a strictly pseudo linear function is essentially a strictly monotoninic function that has a non vanishing first deriviative? Ie a strictly increasing pseudolinear function, or a pseudo concave and pseudo convex function that is also strictly quasi convex and strictly quasi concave>
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