# The Banach Contraction Principle (Banach Fixed-Point Theorem)

I wanted to write about the Banach Contraction Principle (from here out the BCP) because of the intuitive mathematical beauty it illustrates. It's proof is elegant, but even more than that, the theorem provides intuition into problems from fields diverse as ODEs and differential geometry.__If $X$ is a complete metric space and $T:X \to X$ is a contraction, then $T$ has a unique fixed point $a \in X$. Futhermore, for any $x \in X$, $a = \lim T^i (x)$.__

**Theorem Statement***Proof:*Before starting, it should be noted that a map $T$ is a

*contraction*if for some constant $K < 1$, we have $d(Tx, Ty) \leq K \dot d(x, y)$ for all $x, y \in X$.

Now, to begin, fix any point $x_0 \in X$ and set $x_i = Tx_{i-1}$. Letting $\delta = d(x_0. x_1)$ we have \begin{align*} d(x_0, x_1) &\leq d(x_0, x_1) + d(x_1, x_2) + \cdots _ d(x_{i-1}, x_i) \\ &\leq d(x_0, x_1) + K \cdot d(x_0, 1) + K^2 d(x_0, x_1) + \cdots \\ &= \delta(1 + K + K^2 + \cdots) \\ &= \delta/(1-K) \end{align*} Note that for $m \geq n$, the following holds: $$d(x_m, x_n) \leq K \cdot d(x_{n-1}, x_{m-1}) \leq K^2 \cdot d(x_{n-2}, x_{m-2}) \leq \cdots \leq K^n \cdot d(x_0, x_{m-n}) \leq \delta K^n/(1-K)$$ which limits to 0 since $K < 1$.

The key insight here is that the $\{x_i\}_{i\in \mathbb{N}}$ form a Cauchy sequence. Since we assumed that $X$ is complete, this sequence converges and we can assign $a = \lim_{\i \to \infty} x_i$. Thus $Ta = \lim T x_i = \lim x_{i+1} = a$.

Now we need only show the uniqueness of $a$. Suppose $b$ was another fixed point. Then we'd have $$d(a, b) = d(Ta, Tb) \leq K \cdot d(a, b)$$ And since $K < 1$, we get $d(a, b) = 0$ implying that $a = b$. $\blacksquare$