Intro to Proof Class: Understanding how to interpret $f \colon [0,1] \to \mathbb{R}$; $I: A \to \mathbb R$; $I(f) = \int_0^1 f(x) dx$

Question

I am currently in a intro to proof class, so far I have had a good understanding of proof techniques but after moving onto functions I have become really confused on how to interpret some basic ideas:

For instance, $A=\lbrace f \colon [0,1] \to \mathbb{R} |f\;\text{continuous}\rbrace$ (added this after, sorry for the confusion)

I read this as our domain is the range of values between 0 and 1. Our codomain is any real number. So, we could make functions such as $f(0) = 2, f(0.1) = 100, f(1) = -3$.

Next $I: A \to \mathbb R$ is given by $I(f) = \int_0^1 f(x) dx$

I read this like our domain is any real number, so $I(f) = \int_0^1 f(x) dx$ means we input a real number we got from $A$, and $f(x)$ is the real number's integral?

So, an example would be $f(0) = 2$, so $I(2) = \int_0^1 f(2) dx$ = $2$?

I ask this because I want to prove surjection and disprove injection, but I am so lost on what these functions CAN and CANNOT do that I can't move onto disproving or proving the statement.

Any advice would be extremely helpful.

Answer 1

Let's start with elaborating what $A$ is.

$A$ is a set that consists of all continuous functions $f: [0,1] \to \mathbb{R}$. For example $f(x) = x^2$ is in $A$, but

$$ f(x)=\begin{cases} -1, & x<0.5 \\ +1, & x\geq 0.5 \end{cases} $$

is not, because the limit in $0.5$ doesn't exist. So basically $A$ is a collection of all continuous function on given interval.

Now you define $I: A \to \mathbb{R}$, that is often called as a "functional", i.e., a mapping from some set to $\mathbb{R}$. You take an element of your set and send it to some real number via $I$. In your case, you take a function and send it to its definite integral on given interval.

About surjection, let's notice that if $f$ is a constant function(e.g. $f(x) = 5$, every $x$ is being sent to $5$) such that $f(x)=c$, then $I(f) = c$. So surjectivity holds: for every real number $c$ you have a function $f(x)=c$ such that $I(f)=c$.

About injection, let's notice that obviously, $I(0) = 0$. Can we find such a function $f \in A$ that $I(f) = 0$? Yes, for example let $f = 2x-1$. We have that $I(f) = \int_0^1 f(x)dx = \int_0^1 (2x-1) dx = x^2 - x + C\big\vert_0^1 = 0$. Therefore $i$ is not injective: distinct functions are not being mapped to distinct values.

Answer 2

This is a definite integral. What does a definite integral do? It maps your function to a real number. Think of $I$ as a mapping which takes your function and integrates it from 0 to 1. I'm assuming $A$ is the space of functions $f: [0, 1] \to \mathbb{R}$. Lets say $f(x) = x$ restricted to $[0, 1]$. Then $I(f) = \int_0^1 x dx = \frac{1}{2}$. However, lets say $g(x) =0$. Then $I(g) = \int_0^1 0 dx = 0$

Answer 3

It looks like $A$ is the set of functions from $\mathbb{R}\to\mathbb{R}$, at least to me. I say this because $I$ takes a function as an input and returns a real number as an output. I'm not sure how much analysis you've had up to this point, but the set of functions from $\mathbb{R}\to\mathbb{R}$ is of a greater cardinality than $\mathbb{R}$ itself. That essentially means A is "bigger" than $\mathbb{R}$, so $I$ cannot be injective.

$I$ is definitely surjective though, take $y\in\mathbb{R}$, then take $f\in A$ such that $f(x)=y$, then $I(f)=y$.

EDIT: $A$ is actually the same cardinality as $\mathbb{R}$, but $I$ is still not injective because $f(x)=1$ and $g(x)=2x$ both map to $1$:

$$I(f)=\int_0^11dx=1,\quad \quad I(g)=\int_0^12xdx=1$$