Question about continuous functional calculus and its application

Question

I recently started learning about the topic functional calculus. My problem is that I have no idea on how to use it for, say, solving problems, exercises etc.

Here is a short review of what I learned so far.

The idea behind functional calculus seems to be that one would like to "apply" a function $f$ to an operator $T$. If for example $f: \mathbb{R} \rightarrow \mathbb{R}$ and $T: H_1 \rightarrow H_2$, the term $f(T)$ does not make any sense, since the domain of $f$ is $\mathbb{R}$.

But one could still make sense of the term $f(T)$.

For example, if we consider a matrix $M \in M_{n \times m }$ and $f$ to be a polynomial, for example $f(x)=3x^3-x^2$. Then $f(M)$ could be viewed as $3M^3-M^2$, which are defined for matrices, so everything is fine.

Let $H$ denote a complex Hilbert space. And $L(H)$ denote the set of bounded and linear operators on $H$.

Then the range of $T$ was introduced as $R(T):=\{\langle Tx,x\rangle: \lVert x\rVert =1\}$. It is mentioned that $R(T)$ is bounded, thus $R(T)$ is compact.

Then it is shown that for $T \in L(H)$, $\sigma(T) \subset \overline{R(T)}$. (Where $\sigma(T)$ denotes the spectrum of $T$.)

If T is self-adjoint, then $\sigma(T) \subset [m(T),M(T)]$, where $m(T):=\inf\{\langle Tx,x\rangle :\lVert x\rVert=1\}$ and $M(T):=\sup\{\langle Tx,x\rangle :\lVert x\rVert=1\}$.

After those technicalities, it is mentioned that one wants to define $f(T)$ for $f \in C(\sigma(T))$.

Let $f$ be a polynomial with complex coefficients, i.e. $f(t):=\sum_{k=0}^nc_kt^k$. Then $f(T)$ means the operator $\sum_{k=0}^nc_kT^k$.

One crucial thing seems to be that polynomials are dense in the set of continuous functions.

Let $t$ denote the identity function and $1$ denote the constant function $t \mapsto 1$.

Theorem (Continuous functional calculus) Let T \in L(H) be self-adjoint. Then there exists exactly one map $\Phi: C(\sigma(T)) \rightarrow L(H)$ such that

1. $\Phi(t)=T, \Phi(1)=Id$
2. $\Phi$ is linear, $\Phi(fg)= \phi(f)\phi(g)$ and $\Phi( \overline{f})=\Phi(f)^*$
3. $\Phi$ is continuous

We call $\Phi$ the continuous functional calculus of $T$. We will write $f(T):=\Phi(f)$ for $f\in C(\sigma(T))$.

As mentioned at the beginning, I have no idea on how to approach a problem using (continuous) functional calculus. For example, I tried to find some exercise that should be solvable by using functional calculi, but I do not know how to approach those neither how to use functional calculus.

---

Here are some of the Problems I found:

1. Let $T \in L(H)$ be self-adjoint and $\lambda \in \mathbb{C} \setminus(\sigma(T))$. Then $d(\lambda, \sigma(T))=\lVert (T-\lambda Id)^{-1}\rVert^{-1}$

2. Let $T \in L(H)$ be self-adjoint and f \in C(\sigma(T)). Show that the following are equivalent: (i) $f(T)$ is a positive operator (ii) $f \geq 0$

3. If $T \in L(H)$ is a self-adjoint Operator, then there exist two positive operators $T_1,T_2 \in L(H)$ such that $T=T_1-T_2$ (Are $T_1$ and $T_2$ unique?)

---

I assume that I am missing some crucial idea or point, and that's the reason why I have no clue in how to approach the problems above.

Small Edit: My guess in approaching problems by using functional calculus is to define the function $\Phi$ (that is literally called continuous functional calculus of $T$) and then to try to use some denseness argument to get to the result. But how does one "find" $\Phi$. In the Theorem, it is stated that there exists (exactly one) $\Phi$, but not how to find it.

Answer 1

My guess in approaching problems by using functional calculus is to define the function $\Phi$ (that is literally called continuous functional calculus of $T$) and then to try to use some denseness argument to get to the result. But how does one "find" $\Phi$. In the Theorem, it is stated that there exists (exactly one) $\Phi$, but not how to find it.

This is not the approach. You don't have to find $\Phi$, and you don't even have to care about it. All that it matters is that it exists. The point of functional calculus is to do with operators things you can do with functions. A crucial fact is that $\Phi$ is a $*$-homomorphism, which in this case means that you have things like $(f+g)(T)=f(T)+g(T)$, $(fg)(T)=f(T)g(T)$.

Also, because $\Phi$ is a $*$-isomorphism it is automatically continuous. From $\Phi(t)=T$ and being a $*$-homomorphism you get that $\Phi(p)=p(T)$ for any polynomial. And also, by the continuity, if $f=\lim p_n$ is a uniform limit of polynomials, then $$ f(T)=\Phi(f)=\lim_n\Phi(p_n)=\lim_np_n(T). $$

---

2. Let $T \in L(H)$ be self-adjoint and $f \in C(\sigma(T))$. Show that the following are equivalent: (i) $f(T)$ is a positive operator (ii) $f \geq 0$

Suppose that $f\geq0$. From property 2 of $\Phi$, you know that if $f$ is real then $f(T)$ is selfadjoint. So $f(T)$ is selfadjoint. Because $\Phi$ is a $*$-isomorphism between $C^*(T)$ and $C(\sigma(T))$, you have that $\sigma(f(T))=\sigma(f)=f(\sigma(T))$ (this is known as the Spectral Mapping Theorem). Then $f(T)$ is positive. Now suppose that $f(t)<0$ for some $t\in\sigma(T)$. As $f(t)\in f(\sigma(T))=\sigma(f(T))$, the spectrum of $f(T)$ contains a negative point and so $f(T)$ is not positive.

---

3. If $T \in L(H)$ is a self-adjoint Operator, then there exist two positive operators $T_1,T_2 \in L(H)$ such that $T=T_1-T_2$ (Are $T_1$ and $T_2$ unique?)

If you think of applying this to a real function $f$, you have $f=f^+-f^-$, where $f^+=\max\{f,0\}$ and $f^-=-\min\{f,0\}$. These two functions are continuous, and $f^+(T)$ and $f^=(T)$ are positive by the previous exercise. Then you can take $T_1=f^+(T)$, $T_2=f^-(T)$. The decomposition is clearly not unique, for given any positive operator $S$ you have $T=(T_1+S)-(T_2+S)$. But the ones we found above also have the property that $T_1T_2=0$; with this additional property the decomposition is unique. Indeed, suppose that $T=R-S$ with $R,S\geq0$ and $RS=0$. We have $T^2=R^2+(-S)^2$ and, more generally, $T^n=R^n+(-S)^n$. Taking linear combinations we get that $p(T)=p(R)+p(-S)$ for all $p\in\mathbb C[x]$. And, taking uniform limits, $f(T)=f(R)+f(-S)$ for all $f\in C(\sigma(T))$. As $R\geq0$ we have $f^+(R)=R$ and, as $-S\leq0$, we get $f^+(-S)=0$. Then $$ T_1=f^+(T)=f^+(R)=R, $$ and so $S=T_2$.

---

Here is another example:

Show that $T$ selfadjoint is positive if and only if $T=S^*S$ for some $S$.

If $T=S^*S$ then it is easy to show that $T$ is positive. For the converse, if $T$ is positive let $S=f(T)$, where $f(t)=t^{1/2}$. This works because $f\in C(\sigma(T))$ since $\sigma(T)\subset[0,\infty)$. So $S$ is positive and $S^*S=S^2=(T^{1/2})^2=T$.

---

1. Let $T \in L(H)$ be self-adjoint and $\lambda \in \mathbb{C} \setminus(\sigma(T))$. Then $d(\lambda, \sigma(T))=\lVert (T-\lambda Id)^{-1}\rVert^{-1}$

I left this one for last because it requires normal functional calculus (basically, what you wrote works for normal operators, with the same properties for $\Phi$).

Let $g\in C(\sigma(T-\lambda I))$ be $g(t)=1/t$. Since $gz=1$, (with $z(t)=t$) $$ g(T-\lambda I)\,(T-\lambda I)=g(T-\lambda I)\,z(T-\lambda I)=(gz)(T-\lambda I)=1(T-\lambda I)=I, $$ we have $g(T-\lambda I)=(T-\lambda I)^{-1}$. Because for a normal operator the norm agrees with the spectral radius, we have $\|f(T)\|=\|f\|_\infty$ for all $f\in C(\sigma(T))$. Then \begin{align*} \frac1{\|(T-\lambda I)^{-1}\|} &=\frac1{\|g\|_\infty}=\frac1{\max\{|g(t)|:\ t\in\sigma(T-\lambda I)\}}\\[0.2cm] &=\min\Big\{\frac1{|g(t)|}:\ t\in\sigma(T-\lambda I)\Big\}\\[0.2cm] &=\min\{|t|:\ t\in\sigma(T-\lambda I)\}\\[0.2cm] &=\min\{|t-\lambda|:\ t\in\sigma(T)\}\\[0.2cm] &=d(\lambda,\sigma(T)). \end{align*}