If $X$ is a complete metric space, then a contraction mapping is defined by $T:X\rightarrow X$ where $d(T(x),T(y))\leq cd(x,y)$ where $0\leq c < 1$ and $d$ is the appropriate metric/norm.
I know if we are considering spaces such as $[0,1]$ and the function $f$, it means the function $f$ has a fixed point, namely $f(x)=x$ where $x\in [0,1]$. How should interpret $T:L^p([a,b])\rightarrow L^p([a,b])$ via an operator $T(g)$? Does this mean there is a unique function $g$ in $L^p([a,b])$ where $T(g)=g$? In other words, the operator inputs a function from the domain space and outputs a function from the same space? I am struggling to get a clear intuition for this.
Next, could we prove the Fourier transform $\mathcal{F}:L^2([0,1])\rightarrow L^2([0,1])$ is a contraction mapping (on a suitable subset) via the following argument:
$(1)$ WTS $||\mathcal{F}f-\mathcal{F}g||_{L^2}\leq c||f-g||_{L^2}$
$(2)$ consider the closed unit ball in $L^2$, i.e. $B_{1}=\{f\in L^2([0,1]):||f||_{L^2}\leq 1\}$
Take the Fourier transform $\mathcal{F}:B_1\rightarrow B_1$ as $\mathcal{F}=\int_{0}^{1}f(x)e^{-2\pi i x} dx$ and show this is a contraction on $B_1$ via the definition.
$(3)$ conclude the Fourier transform maps $L^2$ functions to other $L^2$ functions on the unit ball.
Finally, what other interesting operators and contraction mappings can we consider for $L^p$ and other function spaces?
Yes, an operator $T:L^p([a,b])\to L^p([a,b])$ takes a function $f$ and maps it into another function $T(f)$ in the same space. So having a fixed point, would mean that $T(f)=f$, that is, that the function $f$ is mapped into itself. For example, you could take $T(f)(x)=\int_a^x f(t)\,dt$ and so a fixed point for $T$ would be a function $f$ such that $f(x)=\int_a^x f(t)dt$ for every $x\in [a,b]$. The Fourier transform is defined for functions in $L^2(\mathbb{R})$ and not $L^2([0,1])$. Parseval identity gives $\Vert\mathcal{F(f)}\Vert_{L^2}=\Vert f\Vert_{L^2}$ and so the Fourier transform cannot be a contraction.