help me with this real analysis problem?

Question

f: R-->R

If f is everywhere differentiable with c = sup{lf'(x)l: x in R}<1
why is there a unique y in R for which f(y) = y
*lf'(x)l: absolute value of f'(x)

show counter-example that f is differentiable with lf'(x)l < 1
for any x in R, f need not have any fixed points in R

Answer 1

First, define a new function g(x)=f(x)-x. f(x) has a fixed point iff ∃x: g(x)=0. By the fundamental theorem of calculus:
g(x) = g(0) + [0, x]∫g'(a) da

Let c be sup(|f'(x)|). Then f'(x)≤c<1, and so g'(a)≤c-1<0. So for x>0:
g(x) < g(0) + [0, x]∫c-1 da = g(0) + x(c-1).
And for x<0:
g(x) > g(0) + [0, x]∫c-1 da = g(0) + x(c-1).

However, [x→∞]lim g(0) + x(c-1) = -∞ (since c-1 is negative), and since g(x) is bounded above by this function for x>0, [x→∞]lim g(x) must also be -∞. Conversely, [x→-∞]lim g(0) + x(c-1) = ∞, and since g(x) is bounded below by this functon for x<0, [x→-∞]lim g(x) is ∞. But this means that g(x) is positive for some negative value of x and negative for some positive value of x, thus by the intermediate value theorem, ∃x: g(x)=0 and therefore f(x) has a fixed point. To see that it's the only one, consider that if g(x) had two zeros then by Rolle's theorem there would exist a point where g'(x) is zero, which contradicts the fact that g'(x) is strictly negative everywhere.

Note that this proof remains valid if c is merely the supremum of f'(x), instead of |f'(x)|.

For the counterexample with |f'(x)|<1 (but not sup(|f'(x)|)<1), consider f(x)={1/x+x if x≥1, 2 if x<1}. This function is everywhere differentiable ([x→1-]lim f(x) = [x→1+]lim f(x) = 2, and [x→1-]lim f'(x) = [x→1+]lim f'(x) = 0), and ∀x, |f'(x)|<1, but f(x) has no fixed points.