I was reading a book and came across the following: If $f$ is a linear functional on a C*-algebra $A$ such that $a\leq f(a)1$ for all $a\geq 0$, then $A$ must be finite dimensional.
One possible way to see this is by a way of contradiction. That is, if $A$ is infinite dimensional, the there must exist a linear functional $f$ such that the above inequality does not hold. I could not come up with any such example or could not see any other way to go about it. Any help would be highly appreciated.
This is definitely using far heavier machinery than necessary. But assume to the contrary that we have such an $f$ defined on an infinite-dimensional $A$. Note that $f$ is clearly positive and therefore necessarily bounded. By passing to the double dual, we may assume $A$ is a von Neumann algebra and $f$ is normal. Any infinite-dimensional von Neumann algebra admits a separable infinite-dimensional abelian von Neumann subalgebra, so by restricting to such a subalgebra, we may assume $A$ is a separable infinite-dimensional abelian von Neumann algebra and $f$ is normal. But the structure of such an algebra is known and we see that there exists a sequence of nonzero projections $p_n$ converging to $0$ in SOT. But then $f(p_n)1 \to 0$ in norm since $f$ is normal, but $p_n$ all have operator norm $1$, contradicting the assumption that $p_n \leq f(p_n)1$ for all $n$.