If we have $\mu_{xy}$, why can we only construct the spectral measure if $\| \mu_{xy} \| \le \| x \| \|y \|$?

Question

Definitions

Representation

Let $X \subset \mathbb{C}^N$ and $\mathcal{A}$ be an algebra in $\mathcal{C}(X)$. Also, we denote $L(H)$ as the set of all linear operators on HIlbert space $H$.

We call $\Phi : \mathcal{A} \ni u \rightarrow \Phi(u) \in L(H)$, if:

1. $\Phi$ is linear and multiplicative
2. $\Phi(e) = I_H$, where $e$ is the neutral multiplicative element of the algebra $\mathcal{A}$
3. There exists a $K > 0$ such, that $\| \Phi(u) \| \le K \|u \|$ for all $u \in \mathcal{A}$

Measure $\mu_{xy}$

A measure $\mu_{xy}$ is called the "elementary measure for $\Phi$", if for any $x,y \in H$:

$$\langle \Phi(u) x, y \rangle = \int_X u(z) d\mu_{xy}$$ for $\| \mu_{xy} \| \le K \| x \| \|y \|$.

Spectral measure

On a Borel set $\mathcal{B}$ in $\Omega$, we call $E : \mathcal{B} \ni \Delta \rightarrow E(\Delta) B(H)$ a spectral measure, if:

1. $E(\Omega) = I_H$, $E(\varnothing) = 0$
2. $E(\Delta)$ is an orthogonal projection in $H$ for all $\Delta$
3. $E(\bigcup_{n}^{\infty} \Delta_n) = \sum_{n}^{\infty} E(\Delta_n)$ for all disjunctive $\Delta_n$

Important: In this context, we will assume, that $\Phi$ is a natural representation ("functional calculus") of an $N$-tuple of operators $(T_1, \ldots, T_N)$, i.e.

$$\Phi(u) := u(T_1, \ldots, T_n)$$

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Now to the question: If we have $\mu_{xy}$, why can we only construct the spectral measure (for the operator tuple $(T_1, \ldots, T_N)$), if $\| \mu_{xy} \| \le \| x \| \|y \|$ (and thus $K = 1$)?

We know that we then have:

$$\mu_{xy} (\Delta) = \langle E(\Delta) x, y \rangle$$

But why is the assumption of $\| \mu_{xy} \| \le \| x \| \|y \|$ necessary?

We construct it by defining $q_{\Delta} (x,y) := \mu_{xy} (\Delta)$ and then using the Riesz-Frechet theorem (i.e. there exists an operator $A$ such, that $q_{\Delta} (x,y) = \langle A x, y \rangle$). Is the assumption $\| \mu_{xy} \| \le \| x \| \|y \|$ necessary here somehow?