What does $C([0,1))$ stand for in the picture below?

Question

Could anyone tell me what $C([0,1))$ stands for in the picture below?

If I see "$f \in C[0,1)$" anywhere, I'd read it as "the function $f(x)$ is continuous for all $x\in[0,1)$". But this one should be different...

Any help is appreciated.

Answer 1

$C(I)$ is the set (in fact, vector space) of all continuous functions $I\to\Bbb R$. Here, we want to define a map $\Phi\colon U\to V$ where $U=V=C([0,1))$. To do so, we must specify, for each $g\in U$, some $v\in V$, which we then call $\Phi(g)$. Now $\Phi(g)$ (as well as $g$ itself) is a function $[0,1)\to\Bbb R$, so to spell out what $\Phi(g)$ is, we have to say what $\Phi(g)(x)$ is for every $x\in[0,1)$. This is done here: The function $\Phi(g)$ maps $x\in[0,1)$ to $e^{\int_0^xg(t)\,\mathrm dt}\in \Bbb R$. Or, for $g\in C([0,1))$, the function $v:=\Phi(g)\colon [0,1)\to\Bbb R$ is given by $$ v(x)=e^{\int_0^xg(t)\,\mathrm dt}.$$ I suspect that it was the double parentheses $\Phi(g)(x)$ that confused you here ...

Of course, one has to verify that the function $\Phi(g)$ is indeed continuous. But as $g$ is continuous, the integral exists, and it is a continuous function of the upper limit (in fact, this is even continuously differentiable, according to the fundamental theorem). Taking the exponential, doesn't destroy this continuity.