I would like to find a solution of:
\begin{equation} \ln(4x) = x \end{equation}
My idea was to use the Banach contraction theorem: "If $E$ is a closed non empty set in a Banach space $V$ and $T: E \rightarrow E$ a contraction of constant $\alpha$, then $T$ has only one fixed point $\tilde{x}$ and $\forall x \in E: T^n(x) \rightarrow \tilde x$ for $n \rightarrow +\infty$ and $T^n(x) = T \circ T \circ T \circ \dots \circ T$ $n$ times".
Let's work in $E \equiv \mathbb{R} \ \cap \ [1.1, 3]$, with functions $f: E \rightarrow E$ such as $||f(x)||_E \equiv |f(x)| = f(x)$ in our $E$.
We also define $T: E \rightarrow E: x \mapsto T(x) = \ln(4x)$. Here are $y = x$ and $y = T(x)$ drawn below with Geogebra:
We see that $T$ has a fixed point between $x=2$ and $x=3$, but let's prove it:
- $E$ is a non empty closed space;
- $V$ defined by functions from $E$ to $E$ with the norm $||f(x)||_E$ is a Banach space (as $\mathbb{R}$ is a Banach space itself);
- $T(x) \equiv \ln(4x)$ is a contraction: let $x, x' \in E$:
$$||\ln(4x) - \ln(4x')||_E \le \alpha ||x - x'||_E$$ $$\Longleftrightarrow \ln\left(\frac{x}{x'}\right) \le \alpha( x - x')$$
I'm not sure how to prove the end of that part, but as $\frac{\partial \ln(x)}{\partial x} < \frac{\partial x}{\partial x}$ in $E$, I believe this is correct.
Now:
$$T^\infty(x) = \ln(4\ln(4\ln(4\ln(4\ln(4\ln(4 \dots $$
I don't know how to finish this.
Could someone help me finishing this, and occasionnaly help me proving $\ln(4x)$ is contractant please ? Unless it isn't ?