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Currently I'm self studying functional analysis, namely functions of operators. In the text, the author gives the following paragraph:

Paragraph A: First notice that for a self-adjoint operator $A$ and $p_n(t)$ a real polynomial, the operator $p_n(A)$ is also self-adjoint. Therefore, the operator $\varphi(A):=\lim P_n(A)$ is self-adjoint.

Before I get to what I don't understand about Paragraph A, let me give some background in the format of def./thm.:

Definition 1: Let $H$ be a Hilbert space. An operator $A:H\to H$ is called non-negative (and we write $A\geq0$) if and only if $\langle Ax,x \rangle\geq0$ for all $x\in H$. This of course implies that $A$ is self-adjoint. Moreover $A\leq B$ means that

  • both $A$ and $B$ are self-adjoint
  • $B-A\geq0$.

Theorem 1: If $A$ is self adjoint, then $A^{2n}\geq0$. Further, if $A\geq0$ it is also true that $A^{2n+1}\geq0$. Combining these two results, it follows that if $A\geq0$, then for any polynormal $P(\lambda)$ with non-negative coefficients we have $P(A)\geq0$.

Theorem 2: Let $A$ be such that $$ m\cdot I\leq A\leq M\cdot I $$ for some $m,M\in\mathbb{R}$ and let $P$ be a polynomial satisfying $P(z)\geq0$ for all $z\in[m,M]$. Then $P(A)\geq0$.

Corollary 1: If $m\cdot I\leq A\leq M\cdot I$ and $P_1(t),P_2(t)$ are real polynomials with $P_1(t)\leq P_2(t)$ for all $t\in[m,M]$, then $P_1(A)\leq P_2(A)$.

Throughout the following post, $a,b,m,M\in\mathbb{R}$ are such that $a<m\leq M<b$, and $A$ is an operator satisfying $m\cdot I\leq A\leq M\cdot I$. Also $K[a,b]$ denotes the set of piecewise continuous bounded functions which are monotone decreasing limits of continuous functions. For a decreasing sequence $\varphi_n$ converging to $\varphi$ we write $\varphi_n\searrow\varphi$.

Lemma 1: Let $\varphi(t)\in K[a,b]$. Then there exists a sequence of polynomial $P_n(t)\searrow\varphi(t)$ as $n\to\infty$ for all $t\in[a,b]$.

This enables us to define an operator $\varphi(A)$ for every $\varphi\in K$:

Definition 2: Let $P_n(t)\searrow\varphi(t)$ for all $t\in[m,M]$. Then the decreasing sequence $P_n(A)\geq P_{n+1}(A)\geq\cdots$ is bounded. So, the strong limit of $\lim P_n(A)$ exists and we call it $\varphi(A)$.

Going back to paragraph Paragraph A, what I don't understand is: any of it. For instance, why does the author insist on speaking of real polynomials $p_n(t)$, when $\varphi(A)$ is defined in terms of $P_n(t)$? It might be important to note here that the polynomials are obtained via the Weierstrass approximation theorem, and, so, might be real polynomials? I'm really not sure.

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2 Answers 2

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The first claim is that if $p_n$ is a real polynomial, and $A$ self-adjoint, then $p_n(A)$ is still-self adjoint. You ask why we need $p_n$ to be real. This is because if $p_n(x) = ix,$ for instance, then $iA$ is no-longer self-adjoint. This is because $\langle x, (iA)y\rangle = -i\langle x,Ay\rangle = -\langle (iA)x, y\rangle,$ where we pick up this extra factor of $-1$ because we need to conjugate the complex number $i$ when moving it from the second coordinate of the inner product to the outside. Generally, if you're reading a math book, you should make sure you know how to prove its claims. If you try writing down a proof that a polynomial of a self-adjoint operator is still self-adjoint, it's easy to see this problem arise.

The second part of the paragraph is because a limit of self-adjoint operators is self-adjoint; this follows quickly from continuity of the inner product.

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  • $\begingroup$ Thanks for the input. You did clear up my first question; however, I still don't see how this shows that $\varphi(A):=\lim P_n(t)$ is self-adjoint. Since to show such, we have to show that $\langle \lim P_n(t)x,x \rangle=\langle x,\lim P_n(t)x \rangle$, which requires self-ajointness of $P_n(t)$, not $p_n(t)$. $\endgroup$ Commented Jul 11, 2021 at 1:15
  • $\begingroup$ @D.Math I think you mean $P_n(A)x,$ not $P_n(t)x.$ But yes, the $P_n(A)$ are self-adjoint, and so $(P_n(A)x, y) = (x, P_n(A)y)$ for each $n.$ Letting $n$ go to infinity, the left side tends to $(\phi(A)x, y)$ and the right to $(x, \phi(A)y).$ $\endgroup$
    – user147556
    Commented Jul 11, 2021 at 1:17
  • $\begingroup$ Ah yes I meant $P_n(A)$...but I don't see exactly where the self-adjointness of the $P_n(A)$ is coming from? $\endgroup$ Commented Jul 11, 2021 at 1:19
  • $\begingroup$ @D.Math $P_n$ is a polynomial, $A$ is self-adjoint. By the first sentence, $P_n(A)$ is therefore self-adjoint. $\endgroup$
    – user147556
    Commented Jul 11, 2021 at 1:25
  • $\begingroup$ I though we needed that the polynomials $P_n(t)$ are real to ensure that $P_n(A)$ is self-adjoint, correct? I guess I'm not seeing where/why $P_n(t)$ are real polynomials. $\endgroup$ Commented Jul 11, 2021 at 1:29
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First, note that the sum of self-adjoint operators is self-adjoint. This can be shown easily.

Second, if $A$ is self-adjoint, then so is $A^n$ for all $n$. This follows by induction on $n$.

Base case: it's trivial that $\langle Ix, y \rangle = \langle x, y \rangle = \langle x, Iy \rangle$.

Inductive step: suppose true for $n$. Then $\langle A^{n + 1} x, y \rangle = \langle A^n Ax, y \rangle = \langle Ax, A^n y \rangle = \langle x, A A^n y \rangle = \langle x, A^{n + 1} y \rangle$ for all $x, y$. So $A^{n + 1}$ is also self-adjoint.

Third, if $A$ is self-adjoint, so is $cA$ for all $c \in \mathbb{R}$. This is also trivial.

The combination of these three facts mean that if $A$ is self-adjoint then so is $p(A)$ for any polynomial $p \in \mathbb{R}[X]$.

The crucial thing here is that we must use polynomials because they're (1) defined for operators, unlike more general functions $\mathbb{R} \to \mathbb{R}$ and (2) preserve self-adjointness.

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