Consider the following fragment from Murphy's book "$C^*$-algebras and operator theory":
Why is the marked line true? I want to know why $\Vert \varphi(a) \Vert \leq \Vert\varphi_K(a)\Vert$. I tried the following. Given $\epsilon > 0$, choose $x \in H$ with $\Vert x\Vert \leq 1$ and $\|\varphi(a)\|^2 \leq \epsilon + \|\varphi(a)x\Vert^2 $. Write $x = y+z$ where $y \in K$ , $z \in K^\perp$. Then $$\|\varphi(a)\|^2 \leq \epsilon + \|\varphi(a)x\Vert^2 = \epsilon + \Vert \varphi(a)y \Vert^2 + \Vert \varphi(a)z\Vert^2 \leq \epsilon + \|\varphi_K(a) \Vert^2 \Vert y\Vert^2 + \Vert \varphi_{K^\perp}(a)\Vert \Vert z \Vert^2$$
If I can show that $\Vert \varphi_K (a) \Vert = \Vert\varphi_{K^\perp}(a) \Vert$, I will be done, but I'm not sure this even holds.
I'll just suppose that $B \subseteq B(H)$ is a C*-subalgebra (so take $B = \phi(A)$) and show that $\|b\| = \|b_K\|$, where $K = [BH]$ and $b_K: K \to K$ is the induced operator, for $b \in B$. Now notice that for $\xi \in K^\perp$ and $\eta \in H$, we have $$ \langle b\xi,\eta \rangle = \langle \xi, b^*\eta \rangle = 0 $$ since $b^*\eta \in K$. Note that this shows that $b|_{K^{\perp}} = 0$ for $b \in B$. So decomposing the Hilbert space as $H = K \oplus K^\perp$, we can write any element $b \in B$ as a matrix with respect to this decomposition: $$ b = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}, $$ where $b_{11}: K \to K, b_{12}: K^\perp \to K, b_{21}: K \to K^\perp, b_{22}: K^\perp \to K^\perp$. As per the above we have $b|_{K^\perp} = 0$, so that $b_{12} = 0, b_{22} = 0$. Moreover, $K$ is $B$-invariant, so $b_{21} = 0$. Consequently, $$ b = \begin{pmatrix} b_{11} & 0 \\ 0 & 0 \end{pmatrix}. $$ Taking norms, we have that $\|b\| = \|b_{11}\|$, which actually gives the desired result since $b_{11} = b_K$.