Let $\mathcal{A}$ be a $C^*$ algebra and suppose that $a,b \in \mathcal{A}$ are self-adjoint and commuting elements that satisfy $$a \leq b \ .$$ If $f$ is an increasing continuous function on $\mathbb{R}$, then I would like to show that $$ f(a) \leq f(b) \ .$$ I do not know how to even start solving this problem. A naive idea would be to approximate the continuous function $f$ by a sequence of monotone increasing polynomials $\{ p_n \}_{n \in \mathbb{N}}$, then I would like to attempt to show that for any monotone increasing polynomial, we have $$ p_n(a) \leq p_n(b) \ .$$ Now, somehow applying the uniform convergence in the context of positive elements of the algebra could give me my result.

One thing that I considered was that since $a$ and $b$ are commuting and self-adjoint, the algebra generated by these two elements and the identity will be an abelian $C^*$-algebra. However, I do not really see how to utilize this fact here.

I guess my biggest problem is that I don't really know how to manipulate the spectrum of $f(b) - f(a)$ and I don't really have any intuition as to why I would expect this result to be true in such generality.

Any help would be greatly appreciated, I would prefer hints to full solutions, or sketches to proofs which I could fill in myself.


Hint: Let $B$ be the subalgebra generated by $a$ and $b$. Since, as you note, it is abelian, it is isomorphic to $C(X)$ for some space $X$. Moreover, the continuous functional calculus on an algebra of the form $C(X)$ is just given by composition: if $a$ corresponds to the function $g:X\to\mathbb{R}$, then $f(a)$ corresponds to the function $f\circ g:X\to\mathbb{R}$ (if you're not familiar with that fact, prove it!). When interpreted in terms of functions on $X$, then, your statement is easy to prove.

  • $\begingroup$ Thank you for your answer. I am a bit confused on the functional calculus when we have two elements instead of one. If we have a single element $a$ then the functional calculus is a isometric $*$-isomorphic mapping $\rho : C(\sigma(a)) \to C^*(a)$ where we have defined $\rho(f) \equiv f(a)$. If we have two elements, then do we not have two maps? So, $\rho_a : C(\sigma(a)) \to C^* (a,b)$ and $\rho_b : C(\sigma(b)) \to C^*(a,b)$, which are both described as before. $\endgroup$ – Kayle of the Creeks Aug 25 '17 at 21:42
  • $\begingroup$ Right. But now the key point is that $C^*(a,b)$ is actually just $C(X)$, and so you can think about the definition of $\rho_a$ (in the case that your algebra is $C(X)$) and prove that it actually just sends a function $f$ to the composition of $f$ with the function $a$. $\endgroup$ – Eric Wofsey Aug 25 '17 at 21:54

Thanks to the hints by Eric Wofsey, I was able to solve this problem.

Here is a sketch of the full solution for those who also search for this question.

Let $\mathcal{A}$ be a $C^*$-algebra, and consider the $C^*$-algebra which is generated by the normal elements $a_1,...,a_n \in \mathcal{A}$. Denote this $C^*$-algebra by $C^*(a_1,...,a_n)$. In a similar fashion to the case where we have a single generated $C^*$-algebra, there exists an isometric *-isomorphism between $C^*(a_1,..,a_n)$ and the space $C(K)$ where $K \subset \mathbb{C}^n$ is a compact subset. Denote this isomorphism by $\rho$ and note that $\rho$ also has the following property $\rho(a_i) = z_i$ where $z_i(\lambda_1,..., \lambda_n) = \lambda_i$.

We trivially have $C^*(a_i) \subset C^*(a_1,...,a_n)$, and since there is an isometric $*$-isomorphism between $C^*(a_i)$ and $C(\sigma(a_i))$, any continuous function $f \in C(\sigma(a))$ can also be considered a continuous function in $C(K)$.

With these ingredients, we are done. We can compile the steps of the proof as follows.

First show that since $b - a$ is a positive element of $C^*(a,b)$ then it is also a positive element of $C(K)$, note that here $K \subset \mathbb{R}^2$. Using the properties we earlier stated, we know that the corresponding continuous function to the element $b - a$ is given by $g(x,y) = y - x$. The positivity of $f(x,y)$ clearly implies that $y \geq x$.

Using the continuous and increasing function $f : \mathbb{R} \to \mathbb{R}$, we can construct another continuous function $\tilde{f} : \mathbb{R}^2 \to \mathbb{R}^2$ which is naturally the mapping $(x,y) \mapsto (f(x), f(y))$.

Composing these two functions, we see that $h = g \circ f : K \to \mathbb{R}$ is a continuous function such that $h(x,y) = f(y) - f(x)$. Because $f$ is increasing this is a positive element.

Finally, using the inverse of the initial $*$-isomorphism, we can show that the corresponding element in $C^*(a,b)$ for the function $h$ is in fact $f(b) - f(a)$, and since $h$ was a positive element that maps to $f(b) - f(a)$, then $f(b) - f(a)$ is also a positive element, as was required.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.