Question:
Can functions outside bra-ket notation be distributed into the bra and ket? Since I am uncertain exactly how to ask this question, I include the problem below. Essentially, I don't know how eq. (2-55) results from eq. (2-54). Can $\phi_0$ be distributed into the bra and ket in eq. (2-54)?
Edit:
Per @Tortar's comment regarding typos, I have attached an image of the problem from the book in lieu of my LaTeX. Upon further review, the PDF to which @Tortar is referring may be a different edition of the aforementioned book, which is where I assume the typo discrepancy arises!
What I have Tried:
My first thought was to distribute $\phi_1$ in eq. (2-54) to the $f_1$ ket and the $\phi_0$ bra; however, this would yield a strange equation that doesn't look like it would lead to the solution.
\begin{equation} \phi_{1} = N_1^{1/2}(f_1 - <\phi_0\phi_1 | f_1\phi_1>) \tag{1} \end{equation}
Revisiting the definition of the inner product (from here), though, the distribution idea seems shaky at best. Given a function $f(x)$ and $g(x)$, the inner product is defined below.
\begin{equation} <f|g> = \int f(x)^*g(x) \, dx \tag{2} \end{equation}
I think this means that if I choose to "distribute" $\phi_1$ as in eq. (1), the result is shown below:
\begin{equation} <\phi_0\phi_1 | f_1\phi_1> = \int (\phi_0\phi_1)^*(f_1\phi_1)\, dx \tag{3} \end{equation}
Eq. (3) does not advance me toward the solution shown in eq. (2-55).
One thing thing that I think might be useful, but I am not sure how to use it, is the fact that given $z \in \mathbb{C}$, then $|z|^{2} = (\sqrt{zz^*})^2 = zz^*$. The last term in eq. (2-55) has the $|z|^{2}$ form, which is why I bring this up.
You can't distribute $\phi_1$ inside the inner product because $\phi_1$ is an element of a vector space instead the inner product $\langle\phi_0 | f_1\rangle$ is a scalar. The problem with your definition of inner product $\int f(x) g(x)$ is that it doesn't return a scalar indeed I think you should use $\int_a^b f(x) g(x)$.
A way to prove the relation in your book :
From $$\phi_{1} = N_1^{-1/2}(f_1 - \langle\phi_0 | f_1\rangle\phi_0) $$
moving the norm on the left and applying the inner product with $\phi_1$ and using the axioms of an inner product and the assumption of your text you obtain that
$$N_1^{1/2} \overbrace{\langle\phi_1| \phi_1 \rangle}^{1} = N_1^{1/2} = \langle (f_1 - \langle\phi_0 | f_1\rangle\phi_0)| \phi_1\rangle =\langle f_1| \phi_1 \rangle - \langle\phi_0 | f_1\rangle \overbrace{\langle\phi_0 | \phi_1 \rangle}^{0} = \langle f_1| \phi_1 \rangle $$
substituting in $N_1^{1/2} = \langle f_1| \phi_1 \rangle $ the identities
$\phi_1 = N_1^{-1/2}(c\phi_0 + f_1)$ and $c = -\langle\phi_0|f_1\rangle$ gives you
$$N_1^{1/2} = \langle f_1|N_1^{-1/2}(-\langle\phi_0|f_1\rangle\phi_0 + f_1) \rangle \implies$$
$$ N_1 = \langle f_1| f_1 \rangle - \langle\phi_0|f_1\rangle\ \langle f_1|\phi_0 \rangle = \langle f_1| f_1 \rangle - |\langle\phi_0|f_1\rangle|^2 $$