# What does it mean to have a map defined on the continuous functions on the spectrum of an operator - $\phi_A:C(\sigma(A)) \to \mathcal{L}(H)$?

I am reading a book (Reed and Simon vol I) on the continuous functional calculus and it speaks of a map $$\phi_A:C(\sigma(A)) \to \mathcal{L}(H)$$ where$A$ is a self-adjoint operator on a Hilbert space $A$. In fact $\phi_A(P) = P(A)$ where $P(A) = \sum_{n=0}^\infty a_n A^n$. It then says that $\phi_A$ has a unique linera extension to the closure of the polynomials in $C(\sigma(A))$ and this closure is all of $C(\sigma(A))$.

But I don't understand how we are speaking of continuous functions on the spectrum of the operator $A$. What if its spectrum $\sigma(A)$ features discrete real numbers or gaps - how can we we have continuous functions defined on a set that may consist of discrete points or be the union of disjoint intervals with gaps?

• $C(\sigma(A))$ is the set of all complex valued continuous functions on $\sigma(A)$ which is a topological space with respect to the induced topology from $\mathbb{C}$ (or $\mathbb{R}$, since $A$ is self-adjoint). – Lorenzo Quarisa May 22 '18 at 13:23

If $\sigma(A)$ is discrete, then every function is continuous. You can always take the "subspace topology": make the topology of $\sigma(A)$ be $\{\sigma(A)\cap V:\ V\ \text{ is open}\}$. This works fine and gives a natural topology. And one can discuss continuity in any topology, it doesn't matter if it has "gaps". The only thing that changes is that it is now not very intuitive what it means to be continuous.
• So is the subspace topology the topology that is always (as in 99% of the time) used whenever we define continuous functions on the spectrum of an operator? They don't define a topology as far as I can see when they say '$C(\sigma(A))$' so I want to be sure I'm interpreting it correctly? – eurocoder May 22 '18 at 14:16