To make this question relatively self-contained, this post is quite long, but the question itself is rather short.
Consider the following fragments in Takesaki's second volume "Theory of operator algebras II" (section VII.2 "Left Hilbert algebras and weights"):
Why is the red boxed equality true? And if it is not true, is there a way to fix the proof?
Thanks in advance for taking the time to help me!
In order for the answer to be more self-contained, note that we are in the following general setting. We are given a von Neumann algebra $M$ (namely $\mathcal{M}'$) and a left ideal $B \subset M$ (namely $\mathfrak{N}_r \subset\mathcal{M}'$) such that $B \cap B^*$ is a dense $*$-subalgebra of $M$. We define the $*$-subalgebra $A \subset M$ as $$A = B^* B = \Bigl\{ \sum_{i=1}^n x_i^* y_i \Bigm| n \in \mathbb{N}, x_i,y_i \in B \Bigr\} \; .$$ We write $A^+ = A \cap M^+$. Takesaki's Lemma 2.1 shows that $A^+$ is a hereditary subcone of $M^+$ and that $$B = \bigl\{ a \in M \bigm| a^* a \in A \bigr\} \; .$$ We are given a linear functional $\omega : A \to \mathbb{C}$ such that $\omega(a) \geq 0$ for all $a \in A^+$ and $$\sup \bigl\{ \omega(a) \bigm| a \in A^+ \; , \; \|a\| < 1 \bigr\} = \lambda < + \infty \; .$$ We prove that $|\omega(a)|^2 \leq \lambda \, \omega(a^* a)$ for all $a \in A$, which in particular shows that $\|\omega\| \leq \lambda$, as was asked in the posted question.
Claim: for every $a \in A$, there exists a sequence $e_n \in A^+$ such that $\|e_n\| < 1$ for all $n$ and $\omega(e_n^{1/2} a) \to \omega(a)$.
To prove this claim, write $a = \sum_{i=1}^k x_i^* y_i$ with $x_i,y_i \in B$. Put $b = \sum_{i=1}^k x_i^* x_i$. Define $e_n = b(b+1/n)^{-1}$. Since $e_n \leq n b$, we get that $e_n \in A^+$. Clearly, $\|e_n\| < 1$. We have $e_n \leq e_n^{1/2} \leq 1$ and because $e_n$ commutes with $b$, we also find that $$0 \leq b - e_n^{1/2} b \leq b - e_n b = \frac{1}{n} e_n \; .$$ So, for all $n$, we have that $0 \leq \omega(b - e_n^{1/2} b) \leq \omega(b - e_n b) \leq \lambda / n \to 0$.
Denote by $X$, resp. $Y$, the column matrices with entries $x_i$, resp. $y_i$. Then $a = X^* Y$. By the Cauchy-Schwarz inequality, we have $$|\omega(e_n^{1/2} a) - \omega(a)|^2 = |\omega\bigl( (X e_n^{1/2} - X)^* Y \bigr)|^2 \leq \omega\bigl( (X e_n^{1/2} - X)^*(X e_n^{1/2} - X) \bigr) \, \omega(Y^* Y) \; .$$ Then note that $$\omega\bigl( (X e_n^{1/2} - X)^*(X e_n^{1/2} - X) \bigr) = \omega(e_n^{1/2} b e_n^{1/2}) + \omega(b) - \omega(e_n^{1/2} b) - \omega(b e_n^{1/2}) \; ,$$ which converges to zero by the computation above. So the claim is proven.
By the Cauchy-Schwarz inequality, $|\omega(e_n^{1/2} a)|^2 \leq \omega(e_n) \, \omega(a^* a) \leq \lambda \, \omega(a^* a)$. In combination with the claim above, we find the desired inequality $|\omega(a)|^2 \leq \lambda \, \omega(a^* a)$ for all $a \in A$.
