I am self-studying $C^{*}-$ algebra and the material I read tells me that if $C([0,1])$ represents a set of continuous functions $[0,1]\to\mathbb{C}$, then the adjoint on $C([0,1])$ is: $$f\in C([0,1]), f^{*}=\overline{f}$$
I don't get why taking adjoint on $C([0,1])$ is to take the complex conjugate. Any explanation?
Here are three rather immediate explanations.
In a C$^*$-algebra you want elements of the form $f^*f$ to be positive. In $C[0,1]$ you have a natural notion of positivity which is $f\geq0$ if $f(t)\geq0$ for all $t$. So you need $f^*(t)f(t)\geq0$ for all $t$. What recipe can you give me, given a complex number $z$, to choose a number $z^*$ that depends on $z$ and such that it always gives you $z^*z\geq0$? I'm fairly sure you'll come up with $z^*=\overline z$.
It is typical to see $C[0,1]$ as a subalgebra of $L^\infty[0,1]$ and to implement the latter as multiplication operators on $L^2[0,1]$. That is, there is a (very) natural way to see $C[0,1]$ as bounded linear operators $M_f$ on a Hilbert space, with $\|M_f\|=\|f\|_\infty^{\vphantom f}$. It is then an easy exercise to check that $M_f^{*\vphantom f}=M_{\overline f}$.
If instead of $C[0,1]$ you consider $C(X)$ for a finite set $X$, you have $C(X)\simeq\mathbb C^{|X|}$ as vector spaces, as algebras, and as $*$-algebras if you consider $(x_1,\ldots,x_n)^*=(\overline{x_1},\ldots,\overline{x_n})$, which is the natural adjoint in $\mathbb C^{|X|}$. This corresponds again with $f^*=\overline f$.