Question

Let f: [a, b] → [a,b] be a continuous function, where a, b are real numbers with a < b. Show that f has a fixed point (i.e., there exists x e [a, b] such that f(x) = x).

Answer 1

firstly we recall bolzano-theorem and Herval fi [a, b] R Botzano theorem - Let [aeb] be a closed interval and fi Laib be Continuous on [aeb] . if flas, and f(b) are opposite Signs then there exsists a point cin of interval Carb) such that f(C) = 0. Here it is given that fi [a,b] → [aubt is a Continous function. Where cub are real numbers with azb. claim m f has a fixed point Cire. there exsists ze [a,b] Such that f(x)=x). Case 1 - if f(a)= a, on f(b) = 6, then the make existence is proved. So we consider the Second Case Case-2 f(a) a and f(b) = b. Let us consider the function g: [ub] → IR such that 3) = f(z - x) xe Laub] obviously g(x) in continuous on [aeb] Since f(x) and a both are continuous function on [ab]. Now glas = f(aj - a 70 L o rsince flas E [246] 7 6 | Cua ) + Sa fa-a g(b) = f(b)-blo [since f(b) € [a,b] - - and f(b) #b So f(b) <bl. So from ② and ③ we see that g(a) @ and g (b) are opposite sign and g in a continuous function on [a,b ] So by Bolzano -theorem there exsists a boint c in the open interval Coub) such that g(C) = 0

+ f(e)-e = 0 fe) =c for some ce carb) So c in a fixed point of f. Hence combining Case -1 and Case-2 we conclude that f has a fixed point (proved).