what a nice functional equation!
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This problem is screaming Legendre transform at me -- together with equality conditions for equality in the inequality of a convex function and its convex conjugate (Legendre transform). In fact your proof mirrors the proof that the conjugate of the conjugate is the original function.
I've seen so many functional equations but this one was really unique!
The first step to these types of problems feel so obvious once you see it but I find really hard to figure out on my own
I like feeding these problems to ChatGPT to make sure I can still have a job
My favourite videos are about functional equations. I always hope that there exists a non trivial solution to the equation in question. It's a bummer when solution doesn't exist or it's f(x) = x. I try to solve equations by myself before watching a video and it was interesting to recall properties of max. I enjoyed this one a lot.
Do we need to use calculus to maximize { 2xy - y^2 | y ∊ R }? Surely we can write max { 2xy - y^2 | y ∊ R } = max { x^2 - x^2 + 2xy - y^2 | y ∊ R } = max { x^2 - ( x - y )^2 | y ∊ R } = x^2, which establishes f(x) = x^2 as a solution very directly
indeed
how do you prove
max{x² - (x-y)²} = x²
without calculus?
@@adiaphoros6842 the term -(x-y)² is a decreasing function of y, i.e., -(x-y)²
and for the sake of completness, "without calculus" refers to not using derivatives and the like.
You are right sir
I couldn't have figured this out on my own if my life depended on it
What an amazing problem!
Kind of reminds me of the definition of something called a Legendre transform
it's a beautiful equation!!!
Hello Micheal, can you tell me where do you find these type of problems ?
Here's my simple proof:
The beginning is the same as Michael's proof, where we find that f(x) >= x^2 and also that x^2 is a solution.
To prove the uniqueness of this solution, let's assume that another solution exists g(x). We know the inequality g(x) >= x^2, therefore this solution can be written as g(x) = x^2 + h(x) where h(x) is non-negative for all x, and positive for atleast one value of x, call it x=a.
Now we get the following:
x^2 + h(x) = g(x) = max{ 2xy - y^2 - h(y) | y in R}
10:08
First of all I'd like to thank Michel and his team wishing them all a wonderful new year 🔔 🔔 🔔 🔔
And then offer thanks for the content of this video.
It was exhilerating
In general, f = f^* if and only if f = 1/2|| • ||² where f^* is the Fenchel conjugate of f.
Learning more today, and being confused as usual :)
Another way to finish:
Suppose that for some y = a the maximum is achieved, then
f(x) = 2xa - f(a)
رائع جدا كالعادة 🎉
what about f(x)=x^2+c? it achieves the same property if i'm not mistaken
have you tried x^2 plus or minus c ?
Or even x^2 plus or minus a multiple of x and plus or minus constants c as well?
x^2 +/- Ax, x^2 +/- Ax +/-C and of course x^2 +/- C as you noted.
If these are suitable, and there are only countably many families but infinitely many members of each family
Afterthought
Ax^2 Ax^2 +/- C Ax^2 +/- Bx Ax^2 +/- Bx +/- C
Making x^2 a general generator and the other options branches from that generator?
I may be mistaken but it seems like a family of solutions is probable or even families of solutions but ... I have had a busy day.
That doesn't work.
max{2xy - f(y)}
= max{2xy - y² - c}
= max{x² - x² + 2xy - y² - c}
= max{x² - (x - y)² - c}
= x² - c
... which is only equal to f(x) = x² + c when c = 0.
@@zygoloid hmmm - I have not been too diligent on this but ... here goes - based on 4:54
f(x) + f(y) >= 2 xy
Set X := x+c and Y := y := x+c
Then f(X) + f(Y) >= 2XY becomes
(x+c)^2 + (x+c)^2 >= 2(x+c)(x+c) and 2(x+c)^2 >= 2(x+c)^2 as required.
If I can find time tomorrow I'll try a bit more carefully to see where my mistakes may be
EDIT - used another transformation: X=Ax and Y = y =Ax which yields (Ax)^2 + (Ax^2) >= 2(Ax)(Ax) and result as required.
Interim conclusion pending peer acceptance: f(x) = Ax and y=Ax yields a family of solutions for A real.
Another family of solutions is f(X)= X^2 and f(y):= X^2 where wlog X:= x+C yields another family of results for real C
And by extension Ax+C works for real A and C and basis for my claim is
X^2 + Y^2 >= 2XY holds for any of three families of solutions given above using transformations X:= Ax or Ax+C where Y:= X respectively in each case.
The question is formed as a tautology? (pending peer approval) and A or C equal to zero also holds true thus solutions span R2 (or 2-space of reals)
someone please find an error in my reasoning:
f(0) = max{0-f(y)|y∈ℝ} = -min{f(y)|y∈ℝ} = -m
suppose that for some x, f(x)
Do your own homework!! *heeh*
m=0
The contradiction only applies back up to where you made the supposition. You supposed that for some x, f(x) < 0 and showed that that led to a contradiction, so you showed that f(x) >= 0 for all x (which is true).
Note that you showed m is negative within the supposition. Because there was a false premise, that implication is not necessarily true (which it's not, m is zero)
@@n8cantor Agreed - by setting m=0 it defines an event space which holds in 2-space for real numbers and therefore holds for real numbers in n-space for n >= 2 by principles of nested sub-spaces.
And so m also acts as a partitioning in 2-space hence all n-spaces with n >= 2
I don't know what the non-solution envelopes(?) might be called but I suspect the partition of m = 0 is infinitesimally thick and contains infinitely many solutions.
2 male worms, minus, a half worm = 3/2 sand worms.
Yes, that's the kind of math suggested by the radical double induction operator acting as an inverter with possible rational confinement, ready to truncate to rational confinement by shedding the entire opposite sex on the division line, at the border between dominance, and subdominance.
Remember, the postman always rings twice?
gg
Is "super unique" more unique than "unique"?
not a constant or identity function as the solution! yay!
like if you are a moroccan 😂
Is that "if" an implication or an equivalence?
what do you mean exactly?
the problem or the function or what ???
@@ribozyme2899
This is very closely related to the convex conjugate function ( en.wikipedia.org/wiki/Convex_conjugate ). The single-variable version of the convex conjugate is: f^*(x) = sup_y { xy - f(y) } ... it is known that x^2/2 is self-conjugate; that is, f^*(x) = f(x).
legendre Tranform! nYes indeed is is
Thanks for the link. I've been thinking about nested subspaces forced to take common zero.
Plus some simplistic simplifications for example y = mx + c or morphed wlog and probably increased generalities to y + mx + c = 0 {2} or ay + bx = c see {3} P.O. below. Thus a typical 2-space presentation of y = mx + c turns into a 4-space presentation as in {2}. and a line in 2-space has turned into a surface(?) hyperplane (?) "thing" (?) in 4-space and so a "thing" in n-space where n>=4.
Of course it seems by properties of nested subspaces all having a common zero that solving a "relation" (a beautiful relation) creates all manners of wonderful things. For example the solution set in 2-space holds in higher dimension spaces.
Note if Y, M, X and C represent axes in 4 space then y, m, x, and c take values in those axes defined by f(x) and further constraints placed upon it.
By extension and in this case, the solution set partitions every n-space of reals into internal non-solutions, external non-solutions and with solutions forming a partitioning "thing" of single element thickness with infinite solutions. Probably comprising of intersection of all 2-space solution planes?
It was good to see sup and inf and hyperplanes mentioned in your wiki link
Thanks again! RUclips is brilliant! Or as brilliant as the folks using it? 🙂
EDIT
Hmmm an ugh! and ehh! from {2} and see {3} P..O. := ay + bx + c - 0 = empty set/null set and ay + bx +c (or ay + bx = c) is a surface in 5-space that spans 5-space?
Sort of suggests that zero in higher dimensions needs to be considered very very carefully add in a dash of quotients, complexes ... and mix with reals to get a real and imaginary entanglement of stuff?