Spectrum of an element I'm having a little trouble calculating the spectrum of an element: specifically, the element $f(x) = \frac{1}{x}$, as an element of the bounded continuous functions from $[1, \infty)$ with pointwise multiplication and the sup norm.
I'm aware the spectrum of an element is $\sigma(x) = \{ \lambda \in \mathbb{C} : x - \lambda \mathbb{1} $ is not invertible $\}$, (an element is invertible if $\exists\ y\ s.t.\ xy = yx = 1$) and I'm guessing my tired brain is doing something stupid, but by my reckoning is that $\frac{1}{x} - \lambda$ is non-invertible if $\lambda \in (0, 1)$
However, to get that answer, I seem to be doing nothing more than making a wild guess / can't at all prove that this must be the case. Can anyone help me with either the construction of a proof, or give me a reason that this answer is way off and tell me where I'm going wrong?
 A: It's clear that if $\lambda \in (0,1)$, $1/x - \lambda$ is not invertible because its value at $x=1/\lambda$ is $0$.  Next: what about $\lambda = 0$ and $\lambda = 1$.  What could $y$ be in those cases?  Next: what about all other $\lambda$ in the complex  plane?
By the way, since you mention sup norm: I think you must be talking about bounded continuous functions.
A: A few things, first when you are considering $f(x)$ as an operator in must be inside a $C^*$-algebra. You say you are considering it as an elements of continuous functions on $[1, \infty)$ with pointwise multiplication and sup norm, but this isn't such an algebra. There are generally two $C^*$-algebras that one would consider in this case.
i.) $C_0([1, \infty))$ the algebra of continuous function vanishing at $\infty$
ii.) $C_b([1, \infty))$ the algebra of bounded continuous functions.
Note that in order to talk about invertible you need the algebra to be unital so the first one is out. So now I will assume you mean the second.
Recall that the spectrum is a compact subset of $\mathbb{C}$. Now $\lambda\in\sigma(f)\Leftrightarrow f(x)-\lambda$ is not invertible $\Leftrightarrow \frac{1}{f(x)-\lambda}$ is not in $C_b([1, \infty))\Leftrightarrow\frac{1}{\frac{1}{x}-\lambda}\not\in C_b([1, \infty))\Leftrightarrow 0\in \overline{\{\frac{1}{x}-\lambda: x\in[1, \infty)\}}\Leftrightarrow \lambda\in[0, 1]$
