Let $A$ be a unital $C^*$-algebra with a sequence of positive elements $x_n$. It is given that $\lVert x_n\rVert\ge c$ for some $c>0$. Let us define the sequence $X_N=\frac{1}{N}\sum\limits_{n=1}^N x_n$. We claim that $X_N$ cannot converge to zero.
I can show this for positive reals $x_n$. Suppose $\lim X_N=0$. Now, due to given hypothesis, $l=\liminf x_n \ge c$. Given any $\epsilon>0$, there is $k\in\mathbb{N}$ such that $ x_n> l-\epsilon$ for all $n>k$. Then for $N>k$, $$X_N=\frac{1}{N}\sum\limits_{n=1}^k x_n+\frac{1}{N}\sum\limits_{n=k+1}^N x_n>\frac{1}{N}\sum\limits_{n=1}^k x_n+(l-\epsilon)(1-k/N)1$$ Taking limit as $N\to\infty$, we obtain $(l-\epsilon)1\le0$ for all $\epsilon>0$. Therefore, $l=0$, which is a contradiction.
But I cannot prove this for sequence of positive elements in any unital $C^*$-algebra. Can anyone help me in this regard? Thanks for your help in advance.
Edit: In $C^*$-algebra case, let $l=\liminf\lVert x_n\rVert$. Then given $\epsilon>0$, there is $k\in\Bbb{N}$ such that $\lVert x_n\rVert\ge l-\epsilon$ for all $n\ge k$. But it may not imply $x_n\ge (l-\epsilon)1$ for all $n\ge k$ where $1$ is the unit of the $C^*$-algebra. I am unable to apply the same argument.
The space $c$ of convergent sequences is a $C^*$-algebra. Let $\delta_n$ denote the standard basis of $c.$ Then $\delta_n\ge 0,$ $\|\delta_n\|=1$ and $\|X_N\|=N^{-1}.$ Thus $X_N\to 0.$
More generally, if there is a sequence $x_n$ such that $x_n\ge 0,$ $\|x_n\|=1$ and $x_nx_m=0$ for $n\neq m$ then $$X_N^2={1\over N^2}\sum_{n=1}^Nx_n^2$$ Thus $$\|X_N\|^2=\|X_N^2\|\le {1\over N}\to 0 $$ If we replace the condition by $x_n\ge c1$ for a positive constant $c,$ then $X_N\ge c1,$ hence $X_N$ does not tend to $0.$