Inductive/Projective Limits of Topological Algebras It is common to form inductive/projective limits of Banach/Frechet spaces in order to come up with natural topologies for common vector spaces. For instance, 


*

*For $k \ge 0$ and $K_n$ compact increasing with $\bigcup_n K_n =\mathbb{R}$,$$
\text{C}^k(K_1)\subseteq \text{C}^k(K_2) \subseteq \ldots \subseteq \text{C}^k(K_n) \subseteq \ldots \to \text{C}^k(\mathbb{R})$$ is the inductive limit of Banach spaces (an LB-space) where convergence is uniform in all $k$ derivatives on compact subsets. (It is also a Frechet space, but this is due to this explicitly construction, not for LB spaces in general)

*If $K\subseteq \mathbb{R}$ is compact then $$
\text{C}^\infty(K) \leftarrow \ldots \subseteq \text{C}^k(K) \subseteq \ldots \subseteq C^1(K)\subseteq C^0(K)
$$ is the projective limit of Banach spaces (making it Frechet), where convergence is such that each derivative individually converges uniformly on $K$.

*Since we have $\text{C}^k(\mathbb{R})$ well defined for all $k \ge 0$, $$
\text{C}^\infty(\mathbb{R}) \leftarrow \ldots \subseteq \text{C}^k(\mathbb{R}) \subseteq \ldots \subseteq C^1(\mathbb{R})\subseteq C^0(\mathbb{R})
$$ is the projective limit of Frechet spaces (also a Frechet space), where convergence is such that each derivative individually converges uniformly on compact subsets of $\mathbb{R}$.

*For $K\subseteq \mathbb{R}$ compact, we take $\text{C}^k_c(K)$ to be the collection of $k$-continuously differentiable functions on $\mathbb{R}$ with support contained in $K$. Then $$
\text{C}^\infty_c(K) \leftarrow \ldots \subseteq \text{C}^k_c(K) \subseteq \ldots \subseteq C^1_c(K)\subseteq C^0_c(K)
$$ is the projective limit of Banach spaces (making it Frechet), where convergence is such that each derivative individually converges uniformly on $K$.
*We take $K_n$ compact increasing with $\bigcup_n K_n = \mathbb{R}$, $$
\text{C}^\infty_c(K_1)\subseteq \text{C}^\infty_c(K_2) \subseteq \ldots \subseteq \text{C}^\infty_c(K_n) \subseteq \ldots \to \text{C}^\infty_c(\mathbb{R})
$$ which is the inductive limit of Frechet spaces (an LF space), where convergence is such that each derivative individually converges uniformly on compact subsets.


My question is how can the algebraic structures of these spaces be inherited from these constructions? For instance, $\text{C}^k(K)$ has a nice structure as a $C^*$-algebra: can inductive or projective limits inherit this structure in one way or another to form some kind of Banach/Frechet/LB/LF-algebra?
 A: Here is the first part of an exercise I translated for 
one of my students (it is in Bourbaki Théories Spectrales 
Chapter 1 paragraph 2) I am not sure that this book has been 
translated into English by now. 
I have just found the latex file and send it because I think 
that it matches with your problem (too long for a comment though). 
a) Let $A$ be a $\mathbb{C}$-algebra endowed with a locally convex 
topology, show that the following conditions are equivalent: 


* 

* It exists a fundamental system of convex balanced neighbourhoods 
of zero $U_i$ such that $U_iU_i\subset U_i$.

* $A$ is isomorphic with a subalgebra of a product of normed 
algebras. 

* the topology of $A$ can be defined by a family of seminorms 
$p_i$ such that $p_i(xy)\leq p_i(x)p_i(y)$ for all $x,y\in A$. 


If $A$ satisfies (one of) these conditions it is called 
locally $m$-convex. The mapping $(x,y)\mapsto xy$ is then 
continuous (to be shown). If $A$ has a unit and if $G=A^{\times}$ (invertible 
elements in $A$), the map $x\mapsto x^{-1}$ from $G$ to $A$ is 
continuous (to be shown). 
Try it and do not hesitate to interact if needed. (Remark: As each normed 
algebra can be embedded in its completion (2) can be rephrased with Banach algebras). 
Remark With this in view, the algebra of entire functions endowed with the topology 
of compact convergence is $m$-convex. I like the proof that physicists give to explain that it admits no 
norm for the same topology. If it were the case one could consider $a:f\mapsto zf(z)$ (creation operator) 
and $b:f\mapsto \frac{d}{dz}(f)$ (anihilation operator) and, as they are continuous, they would be bounded. 
Then you can check the Heisenberg-Weyl identity $[b,a]=Id$ but it is impossible to get such a pair in a 
(non zero) Banach algebra (consider the action of $ad_b:\varphi\mapsto [b,\varphi]$ on the "coherent state" $e^{ta}$ 
for $t\in \mathbb{C}$ and check $ad_b(e^{ta})=te^{ta}$, this would imply that the spectrum of $ad_b$ is 
the whole $\mathbb{C}$, contrary to the fact that $ad_b$ be bounded).           
