I would be inclined to an "indirect" proof. Suppose g is not equal to h. Then there exist y such that [itex]g(y)\ne h(y)[/itex]. Does there exist x such that f(x)= y?
Given that:
1) f,g,h are real nalued funtions ,with f strictly monotone and the other two monotone
2) g(f(x)) = h(f(x)) =x for all ,x ,belonging to the real Nos
3) the domain of all 3 being the real Nos
Prove that:
g=h for all ,x, belonging to the real Nos
Sorry, I was hoping you would see that brings to and both bring it back... so when restricted to the image of , both act as inverse functions. So, try to determine properties for the image of . For instance, assume What would be? What about ? You don't need to be onto. In fact, it can be extremely discontinuous, and by looking at the properties of its image, then the properties of its inverse, you should be able to quickly arrive at your conclusion.
Without loss of generality, assume that is strictly monotone increasing. Assume that it is also bounded above and . Therefore, as . Therefore, . Because g is monotone, it must be monotone increasing, so . However, , so it cannot be that is bounded above. By the same argument for the infimum, it cannot be bounded below. Therefore, it is unbounded. If it is continuous, then it is onto.
Assume that is discontinuous at some . Then, consider and . It must be that , so for all values .
Apply the same to h, and you are done. You will likely need to be more careful with notation than I have been, but this is the skeleton of the proof.
Because cannot be everywhere discontinuous (as it would be unbounded on every interval). Therefore, its set of discontinuities must be denumerable (for more on this, you might want to investigate the image of under where is any interval. Try to make every point--or even an uncountable number of points--in the interval discontinuous, and you will find that because it is strictly monotone, its image must be unbounded.)
So, this implies that is continuous between each and every one of its discontinuity points. implies that for some is continuous on the half-open interval . So, continuity on that interval allows you to find points arbitrarily close to such that their image is arbitrarily close to . Applying the same thought process for , implies that .