Since y in R+ is unique for all x in R+ such that x·f(y) + y·f(x) =< 2, this means there exists some function g : R+ -> R+ such that y = g(x) for all x in R+. This means that x·f(g(x)) + f(x)·g(x) =< 2 for all x in R+. In other words, the task can be rephrased as: find all f : R+ -> R+ such that there exists some g : R+ -> R+ such that x·(f°g)(x) + (f·g)(x) =< 2 everywhere. Since x > 0, (f°g)(x) + (f·g)(x)/x =< 2/x. I think this is probably helpful.
Good explanation! Understood everything! Thanks!
👏👌
Since y in R+ is unique for all x in R+ such that x·f(y) + y·f(x) =< 2, this means there exists some function g : R+ -> R+ such that y = g(x) for all x in R+. This means that x·f(g(x)) + f(x)·g(x) =< 2 for all x in R+. In other words, the task can be rephrased as: find all f : R+ -> R+ such that there exists some g : R+ -> R+ such that x·(f°g)(x) + (f·g)(x) =< 2 everywhere. Since x > 0, (f°g)(x) + (f·g)(x)/x =< 2/x. I think this is probably helpful.
This is a good way to reformulate the condition. We took a similar approach in our video. Our y_x is equivalent to your g(x).
👍👍👍