# Brouwer and Banach fixed point for $x\mapsto f(x)= \frac{e^x}{1+e^x}$

Consider the function

$$f:X\rightarrow Y, \quad f(x)= \frac{e^x}{1+e^x}$$

a) Define X and Y such that is guaranteed by Brouwer's fixed point theorem that this function has a fixed point;

b) Find the best Lipschitz constant within that X domain that you have defined;

c) Find the fixed point by the method proposed by the Banach fixed point theorem.

My annotations

$\textbf{a)}$ So $X \in \mathbb{R}$ and $Y \in [0,1]$.

I have to find $f(x)=x$ right? If we define $f: X \in [0,1] \rightarrow Y \in [0,1]$ , this being a continuous function, then there exists at least one $x \in [0,1]$ such that $f(x)=x$

Proof: Define $g: [0, 1] \subset \mathbb{R}$ where $g(x) = x - f(x)$, with effect $g$ is continuous. Note that the image of $f$ is defined in the interval $[0, 1]$ and therefore $0 \leq f (0) \leq 1$ and $0 \leq f (1) \leq 1$, then $g (0) \leq 0$ and $g (1) \geq 0$, that is $g (0) \leq 0 \leq g (1)$. By the intermediate value theorem, there exists $x \in [0, 1]$, such that $g (x) = 0$. From this fact it follows that $f (x) = x$.

$\textbf{b)}$ I'm using the Takayama-Math Economics book, in which he says: In case of $\mathbb{R}$, the Lipschitz constant is $$\left| \frac{df}{dx} \right| \leq k$$ so, $$\left| \frac{df}{dx} \right| = \frac{e^x}{(1+e^x)^2} \leq k$$ when $x=0$ we have $k\geq\frac{1}{4}$, the best Lipschitz constant?

$\textbf{c)}$ I did it in a medieval way, which is not worth putting here, but I got to 0.659046068

I am studying it alone, so it must have several mistakes (beginner). I wonder if I answered right.

Function graph

• Possibly Brouwer fixed point theorem? – Will Jagy Mar 20 '18 at 0:59
• lol, sorry! my bad – beginner Mar 20 '18 at 1:10
• well done now what is your question – Guy Fsone Mar 20 '18 at 1:16

a) For the first question you did a very nice attempt but it is just use the Brouwer fixed point theorem and you did it correctly.

b) Note that for $v>0$ we have

$$0\le (v-1)^2\Longleftrightarrow. 4v \le (v-1)^2+4v = (v+1)^2$$

that is $$\frac{v}{(v+1)^2}\le \frac14$$

now take $v=e^x$. then you get that for every $x\in\Bbb R$ $$\frac{e^x}{(e^x+1)^2}\le \frac14$$

Which leads you to the Banach contraction mapping (Banach fix point) theorem.

The last question is probably related to what you did during the lecture. Otherwise I think it should be the iterated sequence. $x_{n+1} =f(x_n)$

• Why did you do that? $$0\le (v-1)^2\Longleftrightarrow. 4v \le (v-1)^2+4v = (v+1)^2$$ – beginner Mar 20 '18 at 1:32
• @beginner you are asked to find the Lipschitz constant and it is what I am doing for you. update made reload your page – Guy Fsone Mar 20 '18 at 1:36
• I had found this result. Was my reasoning right? – beginner Mar 20 '18 at 2:24
• @beginner yes your answer is alright to me – Guy Fsone Mar 20 '18 at 5:56