# For a self-adjoint operator $A$ and $p_n(t)$ a real polynomial, the operator $p_n(A)$ is also self-adjoint.

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 $$a, and $$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.

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.

• 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 $p_n(t)$. Commented Jul 11, 2021 at 1:15
• @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 $(x, \phi(A)y).$
– user147556
Commented Jul 11, 2021 at 1:17
• 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? Commented Jul 11, 2021 at 1:19
• @D.Math $P_n$ is a polynomial, $A$ is self-adjoint. By the first sentence, $P_n(A)$ is therefore self-adjoint.
– user147556
Commented Jul 11, 2021 at 1:25
• 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. Commented Jul 11, 2021 at 1:29

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.