In the paper "An Example of a Non-Nuclear $C^*$-algebra, which has the Metric Approximation Property", Haagerup proves the following lemma:
Lemma 1 Let $\varphi$ be a positive definite function on a discrete group $\Gamma$. There exists a unique, completely positive operator $M_\varphi: C_\lambda^* (\Gamma) \xrightarrow{} C_\lambda^* (\Gamma)$ such that for all $x \in \Gamma$, $$ M_\varphi \lambda_x = \varphi(x)\lambda_x. $$
I understood his proof, but I am questioning why isn't this immediate in the sense that for all $x \in \Gamma$ and $f \in \ell^2(\Gamma)$,
$$ \| \varphi(x)\lambda_x f \|_2 \leq \|\varphi\|_\infty \| f \|_2 = \varphi(1) \| f \|_2 $$ $$ \implies \| \varphi(x)\lambda_x \| \leq \varphi(1)\|\lambda_x\|. $$ Why doesn't this fact imply immediately that the operator defined on $\{\lambda_x :x \in \Gamma\}$ extends to a bounded operator on $C_\lambda^* (\Gamma)$ such that $$ \|M_\varphi \| \leq \varphi(1)? $$
Asking this because Haagerup's proof, although relatively easy, is quite more elaborate than this.
Thanks in advance!