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What does $\mathbb{R}^{[0,1]}$ stand for in the following expression? $\Phi: C[0,1] \to \mathbb{R}^{[0,1]}$, where $C$ is the space of all the continuous function in $[0,1]$ and $\Phi$ is an operator.

EDIT 1

Considering the notation $\mathbb{R}^{[0,1]}$ that means the collection of functions from $[0,1] \to \mathbb{R}$.

If,

$$\Phi: C[0,1] \to \mathbb{R}$$

is the collection of continuous functions from $[0,1] \to \mathbb{R}$.

What is the meaning of this one?

$$\Phi: C[0,1] \to \mathbb{R}^{[0,1]}$$

user3889486
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    Note that $C[0,1]$ is a proper subset of $\mathbb{R}^{[0,1]}$. Your mapping $\Phi$ has not been defined, but one such operator is inclusion. – hardmath Jun 13 '16 at 19:17

2 Answers2

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It denotes the collection of functions $[0, 1] \to \mathbb{R}$. In general, $Y^X$ denotes the collection of functions $X \to Y$.

Regarding your edit, $\Phi : C([0, 1]) \to \mathbb{R}$ is not the collection of continuous functions $[0, 1] \to \mathbb{R}$. It is a function from the set of continuous functions $[0, 1] \to \mathbb{R}$, to the set of real numbers $\mathbb{R}$. So, if $f : [0, 1] \to \mathbb{R}$ is a continuous function (i.e. $f \in C([0, 1])$), then $\Phi(f)$ is a real number (i.e. $\Phi(f) \in \mathbb{R}$).

If instead $\Phi : C([0, 1]) \to \mathbb{R}^{[0, 1]}$, then for any $f \in C([0, 1])$, $\Phi(f)$ is a function $[0, 1] \to \mathbb{R}$.

Michael Albanese
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    +1. For the OP: note the analogy with finite sets. How many functions from a $3$-element set to a $2$-element set are there? $2^3$! EDIT: That's an exclamation point, not a factorial. :P – Noah Schweber Jun 13 '16 at 19:08
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It is the set of all the functions : $[0,1] \to \mathbb{R}$.

Bérénice
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