Originally Posted by

**kathrynmath** Suppose f is a real valued function on (-infinity; infinity): Call x a fixed point of f if f(x) = x:

(a) If f is differentiable and f'(t) is not 1 for every t, prove that f has at most one fixed

point.

Suppose $\displaystyle f(x_i)=x_i\,,\,\,i=1,2\,,\,\,x_1\neq x_2$ , and use the mean value theorem:

$\displaystyle \displaystyle{\exists c\in (x_1,x_2)\,\,s.t.\,\,\frac{f(x_1)-f(x_2)}{x_1-x_2

}=f'(c)$ ...

(b) Show that the function f defined by

f(t) = t + (1 + e^t)^-1

has no fixed point, although 0 < f'(t) < 1 for all t.

(c) However, if there is a constant A < 1 such that |f'(t)| < A for all t, prove that a

fixed point x of f exists, and that x = lim xn, where x1 is an arbitrary number and

xn+1 = f(xn); n = 1; 2; 3; ::::

So you have a sequence $\displaystyle \{x_n\}$ , where $\displaystyle x_{n+1}=f(x_n)$ . Now, this seq. is Cauchy

because [tex]|x_n-x_m|=|f(x_{n-1})-f(x_{m-1})|=|f'(d)||x_{n-1}-x_{m-1}| , using once again

the MVT, and now use that the derivative is bounded.

Thus, there exists $\displaystyle \lim\limits_{n\to\infty}x_n=:x$ . Prove now this x fulfills the claim.

Tonio

b) is the only one where I've really gotten anywhere with

f(t)=t

Then t=t+(1 + e^t)^-1

0=(1 + e^t)^-1

0((1 + e^t)=1

0=1

impossible

no fixed point

a)f'(t)=[(f(x)-f(t))/(x-t)] is not 1

Then f(x)-f(t) is not x-t

c)|f'(t)| < A means there is a max value