Let $Tx = 1+\log(1+e^x)$. Show that $T$ has no fixed points.

Question

Let $Tx = 1+\log(1+e^x)$. Show that $T$ has no fixed points.

This is what I have:

We say that $T$ has a fixed point if $Tx=x$.

$$Tx = 1+\log(1+e^x) = x$$ $$\log(1+e^x) = x-1$$ $$1+e^x = e^{x-1}$$ $$e^{x-1}-e^x - 1 = 0.$$

How do I argue mathematically that this last equation does not have a solution? I see that this can only happen if $$e^{x-1}-e^x = 1$$ but I don't know how to argue that this can't happen.

Thank you for any help and comments in advance.

Answer 1

You are almost there. Factor out $e^x$, you will get an equation of the form $$e^x = \mbox{a negative number}$$.

Answer 2

$e^x$ is an increasing function. Hence, $e^x > e^{x-1}$. Hence, $$1+e^x > 1+e^{x-1} > e^{x-1}$$