kavlas Posted May 3, 2011 Share Posted May 3, 2011 Suppose for the real valued funtions f,g,h ,where f is strictly monotone and the other two monotone the following relation holds: g(f(x)) =h(f(x)) =x for all ,x, belonging to the real Nos. Can we rove that : g=h for all ,x, belonging to the real Nos?? The domain of all 3 being the set of real Nos Link to comment Share on other sites More sharing options...

DrRocket Posted May 4, 2011 Share Posted May 4, 2011 Suppose for the real valued funtions f,g,h ,where f is strictly monotone and the other two monotone the following relation holds: g(f(x)) =h(f(x)) =x for all ,x, belonging to the real Nos. Can we rove that : g=h for all ,x, belonging to the real Nos?? The domain of all 3 being the set of real Nos This problem is not so difficult. If it is not a homework problem, it should be one. What have vyou done so far in attempting to solve it ? Link to comment Share on other sites More sharing options...

kavlas Posted May 4, 2011 Author Share Posted May 4, 2011 This problem is not so difficult. If it is not a homework problem, it should be one. What have vyou done so far in attempting to solve it ? The problem appeared in Gazeta Matematica and the difficulty of the problem is in proving that f is onto . Once this has been proved then the solution is easy since f is one to one . This i have dried so far but in vain Link to comment Share on other sites More sharing options...

DrRocket Posted May 5, 2011 Share Posted May 5, 2011 The problem appeared in Gazeta Matematica and the difficulty of the problem is in proving that f is onto . Once this has been proved then the solution is easy since f is one to one . This i have dried so far but in vain OK you have correctly identified the crux of the matter. Here is a sketch of a proof. I will let you fill in the details. f is strictly increasing therefore injective. g and h must agree on the image of f. Since f has a monotone left inverse (g or h) f cannot be bounded below or above. So either f is surjective or else it is not continuous. If f is surjective we are done since g and h agree on the image of f. Since f is strictly increasing, left-hand and right-hand limits exist everywhere and any point of discontinuity of f is a "jump discontinuity". Since g and h are non-decreasing they must agree in the gap corresponding to any jump discontinuity of f. (Think about it graphically then translate into an analytical argument.) Link to comment Share on other sites More sharing options...

kavlas Posted May 10, 2011 Author Share Posted May 10, 2011 OK you have correctly identified the crux of the matter. Here is a sketch of a proof. I will let you fill in the details. f is strictly increasing therefore injective. g and h must agree on the image of f. Since f has a monotone left inverse (g or h) f cannot be bounded below or above. So either f is surjective or else it is not continuous. If f is surjective we are done since g and h agree on the image of f. Since f is strictly increasing, left-hand and right-hand limits exist everywhere and any point of discontinuity of f is a "jump discontinuity". Since g and h are non-decreasing they must agree in the gap corresponding to any jump discontinuity of f. (Think about it graphically then translate into an analytical argument.) Can the above solution be found in any known mathematical book?? Link to comment Share on other sites More sharing options...

DrRocket Posted May 11, 2011 Share Posted May 11, 2011 Can the above solution be found in any known mathematical book?? Not that I know of for certain as the theorem that you state is a bit unusual. The ingredients can be found in Rudin's Principles of mathematical Analysis. You might also look in Folland's Real Analysis: Modern Techniques and Their Applications but I am not terribly familiar with this one. Why would you care ? Link to comment Share on other sites More sharing options...

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