# Characters of algebra of Schwartz functions

Consider the (non-unital) $$\mathbb{C}$$-algebra (point-wise multiplication) of $$\mathcal{S}$$ of Schwartz functions on $$\mathbb{R}$$.

Question: Does there exist some character (non-zero multiplicative functional to $$\mathbb{C}$$) $$\omega$$ of $$\mathcal{S}$$ that is not the evaluation map at any point in $$\mathbb{R}$$, i.e. does $$\mathcal{S}$$ as an algebra, admit characters other than the functionals given by Dirac measures?

Some remarks:

1. This can not happen if we require $$\omega$$ to be continuous with respect to the $$L^\infty$$ norm (one can extend $$\omega$$ to $$C_0(\mathbb{R})$$, which is a commutative $$C^\ast$$-algebra, then use Gelfand-Naimark), my question does not pose any continuity restriction on $$\omega$$, and can be seen of an algebraic nature.

2. This amounts to the question whether the ideal $$C_c^\infty(\mathbb{R})$$ of compactly supported smooth functions is contained in a codimension $$1$$ ideal of $$\mathcal{S}$$.

• For reference, this old math.SE points to a paper that deals with the case of $C_0(X)$. Jan 30 at 19:19

## 1 Answer

Let $$m$$ be a multiplicative functional. Let $$A={\mathbb C}\oplus\mathcal S$$ be the algebra $$\mathcal S$$ extended by the constant functions. This algebra is unital. Setting $$m(f+\lambda)=m(f)+\lambda$$ extends $$m$$ to a multiplicative functional on $$A$$. For $$f\in A$$ let $$s(f)=\sup_{x\in\mathbb R}|f(x)|$$. Assume $$s(f)<1$$. Then we claim that $$\frac1{1-f}$$ lies in $$A$$. We have $$\frac1{1-f}-1=\frac f{1-f}$$ and any derivative of the latter is of a quotient of a polynomial in the derivatives of $$f$$ divided by a power of $$1-f$$. One concludes that $$\frac f{1-f}$$ lies in $$\mathcal S$$, hence $$1-f$$ is invertible in $$A$$. For $$\lambda\ne 0$$ we have $$f-\lambda=\lambda(\frac f\lambda -1)$$, so if $$|\lambda|>s(f)$$, the latter is invertible. Now we have $$m(f-m(f))=0$$, hence $$f-m(f)$$ is not invertible, hence $$|m(f)|\le s(f)$$. This means that $$m$$ is continuous in the sup-norm. Hence it extends to a continuous functional on $$C_c({\mathbb R})$$, i.e., a Radon measure. For this to be multiplicative, it must be a point measure.