Why are positive linear functionals on $C^*$-algebras always bounded?

Question

We say that a linear functional $f$ on a $C^*$-algebra $A$ is positive if $f(a^*a)\geq 0$ for all $a\in A$. Why must it be the case that every positive linear functional on a $C^*$-algebra is bounded?

Answer 1

For self-adjoint elements $a$, we have the inequality $-\lVert a\rVert e\le a\le\lVert a\rVert e$, where $e$ is the identity. So if $f$ is a positive linear functional, $-\lVert a\rVert f(e)\le f(a)\le\lVert a\rVert f(e)$ follows; i.e., $\lvert f(a)\rvert\le\lVert a\rVert f(e)$. For non-selfadjoint $a$, write $a=b+ic$ with $b$ and $c$ selfadjoint and use the result just shown.

Answer 2

The following proof can be found in $\S$4.1 of the course notes A (Very) Short Course on $C^*$-Algebras by D.P. Williams.

Let $f$ be a positive linear functional on a $C^*$-algebra $A$. Suppose that $$\{f(a):a\in A^+,\,\|a\|\leq1\}$$ is unbounded. Then there exists a sequence $a_n\in A^+$ with $\|a_n\|\leq1$ and $f(a_n)\geq n$. Letting $a:=\sum_{k=1}^\infty\frac{a_k}{k^2}$ and $b_n:=\sum_{k=1}^n\frac{a_k}{k^2}$ for all $n$, we have that $a-b_n=\sum_{k=n+1}^\infty\frac{a_k}{k^2}\in A^+$ (since $A^+$ is closed under addition and scalar multiplication by non-negative reals, and is norm-closed).

Since $f$ is positive, this gives that $f(a)\geq f(b_n)\geq\sum_{k=1}^n\frac{1}{k}$ for all $n$, a clear contradiction. We conclude that $$m:=\sup\{f(a):a\in A^+,\,\|a\|\leq1\}<\infty.$$ Since every element of a $C^*$-algebra can be expressed in the form $x+iy$, where $x$ and $y$ are self-adjoint and have norm no greater than $\|a\|$, and each self-adjoint element is a difference of positive elements with norm no greater than $\|a\|$, we have that each $a\in A$ with $\|a\|\leq 1$ can be expressed in the form $$a=b_1-b_2+i(b_3-b_4),$$ where each $b_i\in A^+$ has norm no greater than $1$. Then $$|f(a)|\leq f(b_1)+f(b_2)+f(b_3)+f(b_4)\leq 4m$$ shows that $\|f\|\leq 4m<\infty$ and completes the proof.