In Axler's "Linear Algebra Done Right", there is section of examples of subspaces. There is a sentence that goes:
The set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbf{R}^{[0,1]}$
I am having trouble trying to make sense of the description even before proving if that set is a subpace.
Based on these question asked on the forums: What is $\mathbb R^{\mathbb R}$ as a vector space?
What does the vector space R^[0,1] mean?
$\mathbf{R}^{[0,1]}$ corresponds to the set of functions in the interval [0,1], so a vector $f \in \mathbf{R}^{[0,1]}$ could be seen as an infinitely uncountable tuple where $(f(x_0), f(x_1), ….), x_i \in [0, 1], f(x_i) \in \mathbf{R} $
My question is, if a function f is not continuous in $[0,1]$, can it still an element of $\mathbf{R}^{[0,1]}$ ? For example, f(x) = ln(x)? If a function is not continous in 0 for example, then the tuple that represents ln(x) as a vector would have one less element than, for example, $f(x) = x^2$ ?
Best Answer
The functions in ${\mathbb R}^{[0,1]}$ all have domain $[0,1]$, so $\ln(x)$ isn't in there. On the other hand that fact doesn't really have anything to do with continuity. The function $f:[0,1]\to {\mathbb R}$ defined by $$ f(x) = \begin{cases} 0 & x\in{\mathbb R} \setminus {\mathbb Q} \\ 1 & x\in {\mathbb Q} \end{cases} $$ does belong to ${\mathbb R}^{[0,1]}$, even though it is (very) discontinuous. The only requirement for being in ${\mathbb R}^{[0,1]}$ is that you are a function with domain $[0,1]$ whose values are real numbers.