Injective on $L^n$: what does it mean? I came across the following sentence in an article:

$f$ is injective on $L^1(\mathbb{R})$

where $f$ is some function of $g$. Now I have a vague idea that injective= one-to-one, and can visualize it in $L^1$. Does it mean anything to say injective on $L^n$? Can someone illustrate it with an example in $L^2$?
 A: Presumably, it means that $f$ takes distinct functions in $L^1(\mathbb{R})$ to distinct functions (you say $f$ is "some function of $g$", but you don't say who $g$ is).
Even though $L^1(\mathbb{R})$ is all sorts of things (a measure space, an algebra, a ring of functions, a vector space over $\mathbb{R}$, etc), it is still a set of things. And, being a set of things, you can talk about functions whose domain is $L^1(\mathbb{R})$, and you can talk about such functions being one-to-one, surjective, bijective, etc. 
Same is true for $L^n$, which is likewise a set.
For instance: take a finite interval $[a,b]$, and let $\chi_{[a,b]}$ be the characteristic function of $[a,b]$; that is,
$$\chi_{[a,b]}(x) = \left\{\begin{array}{ll}
0 & \mbox{if $x\notin [a,b]$,}\\
1 & \mbox{if $x\in [a,b]$.}
\end{array}\right.$$
Now, I can define a function $\mathcal{F}\colon L^1(\mathbb{R})\to L^1(\mathbb{R})$ by the following rule: if $g\in L^1(\mathbb{R})$, then $\mathcal{F}(g) = g\chi_{[a,b]}$ (product of functions).
If $g\in L^1(\mathbb{R})$, then so is $g\chi_{[a,b]}$, so $\mathcal{F}$ is indeed a function from $L^1(\mathbb{R})$ to $L^1(\mathbb{R})$. (In fact, it is a linear transformation, which you should check for yourself).
However, $\mathcal{F}$ is not injective on $L^1(\mathbb{R})$: if you take an interval $[c,d]$ with $[c,d]\cap[a,b]=\emptyset$, then $\mathcal{F}(\chi_{[c,d]}) = \mathcal{F}(\mathbf{0})$ (where $\mathbf{0}$ is the constant function with value $0$). Since $\chi_{[c,d]}\neq\mathbf{0}$, then $\mathcal{F}$ is not 1-to-1 on $L^1(\mathbb{R})$. 
