In Theorem 1.3, the following $\mathrm{C}^*$-algebra is considered:
$$A:=\left\{f\in C([0,1],M_2(\mathbb{C})):f(0),\,f(1)\text{ diagonal}\right\}.$$
I want to identify a Hilbert space that these these act on. I want to say that via left multiplication, $g\mapsto fg$, the elements of $A$ are bounded operators on:
$$H:=\left\{f\in L^2([0,1],M_2(\mathbb{C})):f(0),\,f(1)\text{ diagonal}\right\}.$$
Define matrix elements: $$f(x)=\left(\begin{array}{cc}f_{11}(x) & f_{12}(x)\\ f_{21}(x) & f_{22}(x)\end{array}\right),$$ I want to suggest that each of the $f_{ij}$ are in $L^2([0,1])$ where the norm is got from, e.g. $$\langle f_{ij},g_{k\ell}\rangle_{L^2([0,1])}=\int_0^1 \overline{f_{ij}(x)}g_{k\ell}(x),$$
and define the inner product on $H$ by:
$$\langle f,g\rangle =\sum_{i,j=1}^2 \langle f_{ij},g_{ij}\rangle_{L^2([0,1])}.$$
Does this actually check out? As in:
Is the representation of $A$ (via left-multiplication) as bounded operators on $H$ faithful?
Or are there some subtleties and confusions that I am missing out on?
Relevant: $A$ is generated by projections $$p(x)=\left(\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right)$$ and $$q(x)=\left(\begin{array}{cc}x & \sqrt{x(1-x)}\\ \sqrt{x(1-x)} & 1-x\end{array}\right)$$
I went with $\pi(A)\subset B(L^2([0,1],\mathbb{C}^2))$ in the end.