# Locally convex space with a *algebra structure

Consider a $*$-algebra $A$ (over the complex numbers $\mathbb{C}$), which is also a locally convex space, say by the separated family of norms $\{ n_i\}_{ i \in I }$. Assume that the $*$ automorphism is continuous with respect to the locally convex topology.

Then we have a natural notion of an element being positive, namely the positive elements are those of the form $a^{*}a$. Now assume that $f: A \rightarrow \mathbb{C}$ is a linear and positive map, i.e. $f(a^*a) \in \mathbb{R}_{\geq 0}$ for all $a \in A$.

Does it follow that $f$ must already be continuous?

More generally, say we have a densely defined linear positive map $f: D \rightarrow \mathbb{C}$, where $D\subset A$ is dense (in the locally convex topology).

Does $f$ extend to a positive map on $A$? If so, is the extension unique?

For the first question, I know that the answer is positive in the case of a $C^*$ algebra, this is actually how I came up with this question. For the second question, I know that if we are again in the special case of a $C^*$ algebra, if $f$ were already known to be continuous, then it would extend uniquely to the whole algebra, see this question.

## Example

I have a specific example in mind: Consider the space $A = C_c(\mathbb{R}^n,\mathbb{C})$ of continuous functions with compact support. Here the family of seminorms is naturally given by $n_K(f) = \sup_{x\in K} |f(x)|$ where $K\subset \mathbb{R}^n$ is any compact subset. Clearly, it is a $*$-algebra with complex conjugation. In this special case, we can use the Riesz-Markov theorem to see that any positive linear functional is given by integration against some Radon measure, hence is indeed continuous. My interest is that, if the answer to the second question is positive, I can interpret a densely defined (say on $\mathcal{S}(\mathbb{R}^n))$ positive linear functional on $C_c(\mathbb{R}^n,\mathbb{C})$ as a measure on $\mathbb{R}^n$.

I cannot answer your first question in general. But the answer to your second question is no. Take $A$ to $\ell^1(\mathbb N)$, with its usual norm, with pointwise addition an multiplication, and with pointwise conjugation as the involution. In this case, the involution is an isometry, and the positive elements are the sequences with non-negative entries. Let $D$ to be $c_{00}(\mathbb N)$, the sequences with finitely many nonzero terms. Define $$f(a_1,a_2,\ldots)=\sum_n na_n.$$ Then $f$ is positive, but it has no extension to $A$.

In your specific example, though, you can apply Riesz-Markov directly. As stated for instance in Rudin's Real and Complex Analysis, any positive linear functional on $C_c(X)$ for a locally compact Hausdorff space is represented by a unique positive measure.

• Did you mean to define $f$ as the sum of those terms? – jgon Jan 10 '18 at 3:47
• Indeed. Thanks! – Martin Argerami Jan 10 '18 at 4:13
• A comment on the specific example I had in mind: in my situation I really only have a densely defined positive functional. The Riesz-Markov theorem yields continuity of a positive linear functional in the special case $A = C_c(\mathbb{R}^n,\mathbb{C})$, but only for everywhere defined functionals. – Max Jan 10 '18 at 11:05
• Great, thank you for your counterexample. So since $f$ is discontinuous (take the sequence in $c_{00}(\mathbb{N})$ which has a $1$ in the $n$-th place and $0$ elsewhere), it cannot have a continuous extension. So it can also not have a positive extension, since $\ell^1$ is a $C^*$-algebra. Nice! – Max Jan 10 '18 at 11:23
• I'll think about your question. Note that $\ell^1$ is not a C$^*$-algebra (the norm does not satisfy that C$^*$-equality), so what I said might be wrong. I also have to think about that. – Martin Argerami Jan 10 '18 at 14:20

Even for normed $*$-algebras it is not true that positive functionals are necessarily continuous: Consider the subalgebra $A$ of the $C^*$-algebra $C([0,1])$ consisting of all polynomials. Then $f:A\to\mathbb C$, $p\mapsto p(2)$ is well-defined, linear, positive and even multiplicative but not continuous with respect to the uniform norm $\|p\|=\sup\{|p(x)|: x\in [0,1]\}$.

• I find this example very interesting. But notice that it is not positive (for the same reasons that it is not bounded). For instance $p(x)=x-x^2$ is positive (on $[0,1]$, so as an element of $C[0,1]$) but $p(2)=-2$. – Martin Argerami Jan 10 '18 at 16:12
• "... positive elements are those of the form $a^*a$". – Jochen Jan 10 '18 at 16:28
• Fair enough, but then you are not using the positivity of $C[0,1]$, so it doesn't make a lot of sense to talk about "subalgebra", nor extensions. – Martin Argerami Jan 10 '18 at 17:53
• Okay, it is probably not a good idea to present $A$ as a subalgebra of $C ([0,1])$. Nevertheless, it is a counterexample. – Jochen Jan 10 '18 at 18:04
• Jochen's example is correct. $p(x)=x-x^2$ is not a sum of $a^*a$'s. – Yurii Savchuk Jan 11 '18 at 11:39