Riemann hypothesis equivalences,Robin inequality,Lagarias criterion, and Riemann hypothesis

Your counter example is completely wrong. You should consider it in this format: f (x)=sin x^2=O (x^2) and g (x)=x^2, then we have lim |O (x^2)|/x^2 <A when x tends to infinity. This means that |sin x^2|<Ax^2 and since A>=1,then the inequality is true. Just differentiating the inequality for sin x^2>0, we find the inequality 2xcosx^2<2Ax, where is true and means that |f'(x)|<Ag'(x) or f'(x)=O (g'(x)) and the result is complete. If you could not be convinced with my arguments, I will not be able to bring additional clarifications. But you should be aware that this paper has passed peer review process in some prestigious journals and reviewed by many distinguished mathematicians. None of them claimed that this proposition is wrong. I respect to your opinion, but I have no additional time to discuss with you. You are able to whether accepting my arguments or not...decision is yours. Have a nice weekend. I will not respond to your other comments.

I do know very whell what the O(.) and o(.) notations mean. But you are wrong. You give 2 examples where your proposition holds, but you ignore my simple counter-example of f(x)=sin x² => f'(x) = 2x cos x which is an unbounded function and does not satisfy f' = O(g') with g'=1. The result being wrong, the proof is obviously also wrong. In particular, you incorrectly assume that the logical negation of " f < A g " is " f >= A g ", which is not the case. The former is false as soon as there is *one point* (in the considered domain) in which it does not hold. For example, " sin < 0 " is wrong, but "sin >= 0" is not true, either.

Thank you for your comment, although you misunderstood mean of the proposition. We mean of f(x)=O(g(x)) is not a composition of two functions as ordinary analysis. If you accurately review the proposition and its proof, you will understand we mean lim |f(x)|/g(x)<A, when x tends to infinity where A is a positive constant. For example let f(x)=x and g(x)=x, then lim |x|/x<A, when x tends to infinity where A>=1. Then |x|<Ax, just differentiating the both sides of the inequality for x>0 gives us 1<A. The other example: suppose f(x)=sinx=O(g(x)) and g(x)=x, we know sinx/x<1 for x>0, then |O(g(x))| /x <A and A>=1, then |O(g(x))|=|sin x|< Ax, differentiating both sides of the inequality for x>0 and Sinx >0 gives us cos x<A, where is true or for sinx <0, we have again -sinx<A. I hope these help you to understand proposition.

Proposition 1 is wrong. It claims that if g >= 0, g' > 0, then f=O(g) => f'=O(g'). But a simple counter example is g(x)=x and f(x)=sin(x²)=O(g) (even o(g)) and g' = 1 > 0 but f'(x) = 2 x cos(x²) is not O(g') = O(1).