Proof of Contraction Mapping for $f\left(x\right)={e}^{{-e}^{-x}}$

[ozkan12]

Please can you prove that $f\left(x\right)={e}^{{-e}^{-x}}$ is contraction mapping on R...Thank you for your attention...

 

[Evgeny.Makarov]

Try proving that $|f'(x)|<1$.

 

[ozkan12]

I tried this but I didnt get something...Dear professor

 

[Evgeny.Makarov]

What did you get for $f'(x)$?

 

[ozkan12]

I get this ${e}^{-(x+{e}^{-x})}$...But how ${e}^{-(x+{e}^{-x})}$ $\le$ 1 ?

 

[Ackbach]

Well, based on our knowledge of the exponential function $e^x$, we know that $e^x \le 1$ precisely when $x\le 0$. Can you prove that $-(x+e^{-x})\le 0?$

Also, with contraction mappings, it's always a good idea to mention which metric is being used. Is it just $d(x,y)=|x-y|?$

Actually, a more fruitful method of showing this might be to let $g(x)=f'(x)$, and maximize $g(x)$ using the usual calculus methods.

 

[ozkan12]

I know $-\left(x+{e}^{-x}\right)$ $\le$0...But for which "x" we get ${e}^{-(x+{e}^{-x})}$ $\le$ 1? $x\in[0,\infty)$ or $x\in[-\infty,0)$...I didnt know this

 

[Euge]

Hi ozkan12,

You want to prove $x + e^{-x} > 0$ for all $x\in \Bbb R$, for then $e^{-(x + e^{-x})} < 1$ for all $x\in \Bbb R$ as desired. Actually, we can prove that $1/e$ is a suitable contraction constant for $f$ by showing that $1/e = \sup\limits_{-\infty < x < \infty} |f'(x)|$. Now the function $g(x) = x + e^{-x}$ has only one critical point at $x = 0$. Furthermore, $g'(x) = 1 - e^{-x}$ is positive for all $x > 0$ and negative for all $x < 0$. By the first derivative test, $g$ has a global minimum at $x = 0$. The minimum value of $g$ is $g(0) = 1$, so

$$\sup_{-\infty < x < \infty} |f'(x)| = e^{-g(0)} = e^{-1},$$

proving that $1/e$ is a contraction constant for $f$.

 

[ozkan12]

Dear professor,

Why we use supremum ?

 

[Evgeny.Makarov]

We use the mean value theorem to relate derivative and contraction. Namely, if $f$ is differentiable on an interval $I$, then for every $x_1,x_2\in I$ there exists an $x_0\in (x_1,x_2)\subseteq I$ such that $f(x_2)-f(x_1)=(x_2-x_1)f'(x_0)$. Thus, if $S=\sup_{x\in I}|f'(x)|$, then $|f(x_0)|\le S$ and
\[
|f(x_2)-f(x_1)|=|x_2-x_1|\cdot|f(x_0)|\le |x_2-x_1|\cdot S.
\]
Therefore, if $S<1$. then $f$ is a contraction on $I$.

 

[ozkan12]

Dear professor,

Why we use minimum value of "g" ?

 

[Ackbach]

Dear professor,

Why we use minimum value of "g" ?

So, from Euge's post, we have $g(x)=x+e^{-x}$. Then the original function's derivative is $f'(x)=e^{-g(x)}$. Now $e^{-x}$ is a monotonically decreasing function. It follows that a maximum of $f'(x)$, which is what we are trying to find, would happen precisely where $g(x)$ has a minimum.

 

[ozkan12]

Dear Ackbach

I opened new thread on forum...Can you see that ? Thank you for your attention :)