Suppose that we have the following objects:
- $ X $ — a locally compact Hausdorff space.
- $ A $ — a $ C^{\ast} $-algebra.
- $ \pi $ and $ \rho $ — commuting representations of, respectively, $ {C_{0}}(X) $ and $ A $ on a Hilbert space $ \mathcal{H} $, i.e., $$ \forall f \in {C_{0}}(X), ~ \forall a \in A: \qquad \pi(f) \circ \rho(a) = \rho(a) \circ \pi(f). $$
Is there a straightforward way to prove the following inequality without having to bring in the entire machinery of $ C^{\ast} $-tensor-product theory? $$ \forall f_{1},\ldots,f_{n} \in {C_{0}}(X), ~ \forall a_{1},\ldots,a_{n} \in A: \qquad \left\| \sum_{i = 1}^{n} \pi(f_{i}) \circ \rho(a_{i}) \right\|_{\mathscr{B}(\mathcal{H})} \leq \sup_{x \in X} \left\| \sum_{i = 1}^{n} {f_{i}}(x) \cdot a_{i} \right\|_{A}. $$ Thank you!