Bounding norms with sums of positive elements Let $\mathcal{A}$ be a $C^{\ast}$-algebra and suppose that $\lbrace x_1,...x_n\rbrace$ are positive elements of $\mathcal{A}$ such that $\|\sum_{i=1}^{n}x_i\|\leq\lambda$, for some $\lambda\in\mathbb{R}$.
Is it true that I can conclude that each $\|x_i\|\leq\lambda$ ?
 A: Represent the algebra in some Hilbert space and let $P$ be positive. I will use the following statements:


*

*For every $\epsilon$ there exists a norm $1$ vector $x$ so that $\langle x,Px\rangle ≥\|P\|-\epsilon$.

*For any $x$ of norm $1$ you have $0≤\langle x,Px\rangle≤\|P\|$.


Now consider $A,B$ positive and let $x$ be a unit vector so that $\langle x,Ax\rangle≥\|A\|-\epsilon$. It follows that
$$\|A+B\|≥\langle x,(A+B)x\rangle ≥\|A\|+\langle x,Bx\rangle -\epsilon≥\|A\|-\epsilon$$
Since this can be done for any $\epsilon$ you have $\|A+B\|≥\|A\|$.
Now let $\|\sum_i x_i\|≤\lambda$, $x_i$ positive elements. It follows $\sum_{j\neq i}x_i$ is positive and $\|\sum_i x_i\|=\|x_i+\sum_{j\neq i}x_j\|≥\|x_i\|$ from the previous consideration. So $\|x_i\|≤\lambda$.
Proof of the statements used:


*

*Since $P$ is positive $\sqrt{P}$ exists and is positive. There exists norm $1$ $x$ so that $\|\sqrt{P} x\|^2≥\|\sqrt{P}\|^2-\epsilon$ for any $\epsilon$. This equation is however:
$$\langle \sqrt P x, \sqrt P x\rangle =\langle x, Px\rangle ≥\|\sqrt{P}\|^2-\epsilon=\|P\|-\epsilon$$

*This is clear.

A: Since $x_1,\ldots,x_n$ are positive, you have 
$$
0\leq x_k\leq \sum_j x_j. 
$$
It follows that $\|x_k\|\leq\|\sum_jx_j\|\leq\lambda$. 
They key step here is that if $0\leq a\leq b$, then $\|a\|\leq\|b\|$. This can be seen in several ways. For instance, if $S(A)$ denotes the state space,
$$
\|a\|=\sup\{f(a):\ f\in S(A)\}\leq\sup\{f(b):\ f\in S(A)\}=\|b\|,
$$
since $f(a)\leq f(b)$ by the positivity of the state. 
