Positive elements in $C^*$-algebras

Question

I'm trying to prove the following, and I'm not sure if the proof is correct?

If $A,B$ are $C^*$-algerbas, and $f$ is a $*$-homomorphism from $A$ onto $B$ then $f(A_+)=B_+$.

Proof: let $a\in A_+$ then $a=x^*x$ for some $x\in A$, thus, $f(a)=f(x^*x)=f(x^*)f(x)=f^*(x)f(x)$ which implies that $f(a)$ is positive, for the other inclusion, if $b\in B_+$ then $b=y^*y$ for some $y\in B$, since $f$ is onto, $y=f(x)$ for some $x\in A$ and we have $b=f^*(x)f(x)=f(x^*x)$ which implies the result.

And one more question, if we have the following if $f_n(\lambda)$ is a defined on a nbh of $\sigma(a)$ and $f_n(\lambda)\to 0$ as $n\to \infty$ does that imple $f_n(a)\to 0$ as $n\to \infty$,


Edit: Let $f_n(\lambda)=\lambda^{\frac{1}{2}}\left( 1-\sqrt{\frac{\lambda}{\lambda+\frac{1}{n}}}\right )$ then why $\|f_n(a)\|\to 0$,thank you.

Answer 1

For your last question, you might want to use the analytic functional calculus as follows: let $\gamma$ be a rectifiable contour in $\mathbb{C}$ enclosing $\sigma(a)$ on which $f_n$ is defined; this exists since you said $f_n$ is defined on a neighborhood of $\sigma(a)$. Then

$$f_n(a) = \frac{1}{2 \pi i} \int_{\gamma} f_n(\zeta) (\zeta 1 - a)^{-1} \, d\zeta$$

Then the result holds if you can apply some dominated convergence theorem. Unfortunately, I do not see why such a result would hold in our case, unless you have some more information about the convergence of $f_n$. Furthermore, it's not clear whether you mean that all the $f_n$ are commonly defined on some neighborhood of $\sigma(a)$. If, instead, $f_n$ is defined on a neighborhood of $\sigma(a)$ that shrinks as $n$ increases, then it is impossible to find a $\gamma$ enclosing $\sigma(a)$ such that $f_n$ is defined on $\gamma$.

Edit: Since you've given an explicit form for $f_n$, then dominated convergence should hold in that case.

Answer 2

So we've seen that your argument for $f(A_+)=B_+$ works.

Now regarding your functions $f_n$. I understand that you apply them to a positive element $a$. Since each $f_n$ is continuous on $[0,+\infty)$, a fortiori on the spectrum of $a$, the continuous functional calculus defines $f_n(a)$ and the spectral mapping theorem says that $$ \mbox{spectrum}(f_n(a))=f_n(\mbox{spectrum}(A)). $$ Given $f_n$, it follows that the spectrum of $f_n(a)$ is contained in $[0,+\infty)$. Moreover, since $f_n$ is real-valued and $a$ is self-adjoint, we have $f_n(a)^*=f_n(a^*)=f_n(a)$. Therefore $f_n(a)$ is also positive. So $\|f_n(a)\|$ is the maximum of $f_n$ on the spectrum of $a$.

Edit: All you have to do now is to prove that $f_n$ converges uniformly to $0$ on every compact $[x,y]$ in $[0,+\infty)$. I've done it for $x>0$ below. I leave it to you for the case $x=0$.

Case $a$ invertible: To show that $f_n(a)$ converges to $0$, it suffices to show that $g_n(a)$ does for $$ g_n(t)=1-\sqrt{\frac{t}{t+1/n}}. $$ By differentiation, one can see that $g_n$ is decreasing on $[0,+\infty)$. Therefore, by spectral mapping $$ \|g_n(a)\|=\max \mbox{spectrum}(g_n(a))=g_n(\min\mbox{spectrum}(a)). $$

If $a$ is invertible, $x=\min\mbox{spectrum}(a)>0$ and $$ \|g_n(a)\|=1-\sqrt{\frac{x}{x+1/n}}\longrightarrow 0. $$ A fortiori $f_n(a)$ tends to $0$.