Positive operators II

Posted on 2026/06/11 by arkajad

This post is a continuation of Positive operators I. The aim here is to explain how the norm of a bounded self–adjoint operator can be expressed in terms of the associated quadratic form. We will first recall the Riesz representation theorem, then relate bounded operators to bounded bilinear (or sesquilinear) forms, and finally prove that for a self–adjoint operator A the operator norm coincides with the supremum of the absolute value of the quadratic form (x,Ax) on the unit sphere. The main result is the Corollary at the end; it will be used later in the construction of the positive square root of a self–adjoint positive operator in a real or complex Hilbert space.

Linear functionals and the Riesz representation theorem

In this first section we recall the Riesz representation theorem and explain briefly why it characterizes Hilbert spaces. Throughout, H will be a real or complex pre–Hilbert space.

If f is a linear functional on H, its norm is defined as

(1)   \begin{equation*} \Vert f\Vert = \sup_{\Vert x\Vert=1}\vert f(x)\vert. \end{equation*}

Exercise 1.  Show that

(2)   \begin{equation*} \Vert f\Vert = \sup_{x\neq 0}\frac{\vert f(x)\vert}{\Vert x\Vert}. \end{equation*}

(Hint: write x = \Vert x\Vert u with \Vert u\Vert=1 and use homogeneity.)

The space of all bounded linear functionals (i.e. those with a finite norm)
is denoted H^* and is called the dual of H.

Proposition 1. ( Riesz representation theorem )
If f is a bounded linear functional on a Hilbert space H, then there exists a unique vector x_f\in H such that, for all x\in H,

(3)   \begin{equation*} f(x) = (x,x_f). \end{equation*}

Moreover,

(4)   \begin{equation*} \Vert x_f\Vert = \Vert f\Vert. \end{equation*}

A pre–Hilbert space is a Hilbert space if and only if the Riesz representation theorem holds for this space. This characterization has the same formulation in the real and complex cases. A proof of the theorem can be found in almost every textbook on functional
analysis or on Hilbert spaces (though not all sources explicitly treat both the real and complex cases together).
Personally I like the formulation and proof in Ref [1], Ch. V.6.1, p. 136.

Bilinear and sesquilinear forms

We now explain how a bounded operator gives rise to a bounded sesquilinear form, and we prove that the norm of this form coincides with the operator norm. This link will be the main tool in the next section.

If A is a linear operator on a pre–Hilbert space H, we can associate with it a sesquilinear form (bilinear in the real case) f_A defined as

(5)   \begin{equation*} f_A(x,y) = (x,Ay). \end{equation*}

For a general sesquilinear form f(x,y) we define its norm as

(6)   \begin{equation*} \Vert f\Vert = \sup_{\Vert x\Vert=1,\, \Vert y\Vert=1}\vert f(x,y)\vert = \sup_{x,y\neq 0}\frac{\vert f(x,y)\vert}{\Vert x\Vert \Vert y \Vert}. \end{equation*}

Exercise 2.
Prove the last equality.
(Hint: scale x and y to unit vectors as in the previous exercise.)

We now prove that the norm of f_A coincides with the operator norm of A.

Proposition 2. Let A be a bounded linear operator in a real or complex pre–Hilbert space H. With f_A as in Eq.~\textup{(5)} we have

(7)   \begin{equation*} \Vert f_A\Vert = \Vert A\Vert. \end{equation*}

Proof.  Using the Cauchy–Bunyakovsky–Schwarz inequality and the definition of \Vert A\Vert we have

(8)   \begin{equation*} \vert f_A(x,y)\vert = \vert(x,Ay)\vert \leq \Vert x\Vert \Vert Ay\Vert \leq \Vert A\Vert \Vert x\Vert \Vert y \Vert, \end{equation*}

which entails

(9)   \begin{equation*} \Vert f_A\Vert\leq \Vert A\Vert. \end{equation*}

We now prove the opposite inequality.
We have

(10)   \begin{equation*} \Vert f_A\Vert = \sup_{x,y\neq 0} \frac{\vert (x,Ay)\vert}{\Vert x\Vert\Vert y\Vert} \geq \sup_{\substack{y\neq 0\\Ay\neq 0}} \frac{\vert (Ay,Ay)\vert}{\Vert Ay\Vert\Vert y\Vert} = \sup_{\substack{y\neq 0\\Ay\neq 0}} \frac{\Vert Ay\Vert^2}{\Vert Ay\Vert\Vert y\Vert}. \end{equation*}

Here we restrict the supremum in two ways.

First, in the supremum over all x,y\neq 0 we can omit those y for which Ay=0, since then (x,Ay)=0 and these pairs do not contribute to the supremum. Second, for the remaining y we restrict x to be of the form x=Ay. Restricting the set over which the supremum is taken can only decrease its value, hence the inequality.
Finally, for such y we have (Ay,Ay)=\Vert Ay\Vert^2\geq 0, so we can remove the absolute value.

Thus

(11)   \begin{equation*} \Vert f_A\Vert \geq \sup_{\substack{y\neq 0\\Ay\neq 0}} \frac{\Vert Ay\Vert}{\Vert y\Vert} = \sup_{y\neq 0}\frac{\Vert Ay\Vert}{\Vert y\Vert} = \Vert A\Vert, \end{equation*}

which is the desired opposite inequality. Together with (9) this proves (7).

The norm of a self–adjoint operator

In this section we relate the operator norm of a bounded self–adjoint operator to the associated quadratic form. We will use the Parallelogram Law and the polarization identity, first in the real, then in the complex case.

Lemma 1 ( Parallelogram Law ). Let H be a real or complex Hilbert space.
Then, for any x,y\in H we have

(12)   \begin{equation*} \Vert x+y\Vert^2+\Vert x-y\Vert^2 = 2\left( \Vert x\Vert^2+\Vert y \Vert^2\right). \end{equation*}

Exercise 3. Verify the Parallelogram Law by expanding the scalar products
and simplifying.

Proposition 3.
Let A be a bounded self–adjoint operator on a (real or complex) pre–Hilbert space H. Then

(13)   \begin{equation*} \Vert A\Vert = \sup_{\Vert x\Vert=1}\vert (x,Ax)\vert. \end{equation*}

Proof. We begin with one inequality, which follows directly from Cauchy–Bunyakovsky–
Schwarz:

(14)   \begin{equation*} \vert(x,Ax)\vert \leq \Vert x\Vert \,\Vert Ax\Vert \leq \Vert x\Vert\,\Vert A\Vert\,\Vert x\Vert = \Vert A\Vert \Vert x\Vert^2. \end{equation*}

For \Vert x\Vert=1 this gives

(15)   \begin{equation*} \sup_{\Vert x\Vert=1}\vert (x,Ax)\vert\leq \Vert A\Vert. \end{equation*}

To obtain the opposite inequality, we use the polarization identity applied to the Hermitian form f_A defined by

(16)   \begin{equation*} f_A(x,y)=(x,Ay). \end{equation*}

Set

(17)   \begin{equation*} m=\sup_{\Vert x\Vert=1}\vert (x,Ax)\vert. \end{equation*}

Then, for any z\in H we have

(18)   \begin{equation*} \vert(z,Az)\vert\leq m\Vert z\Vert^2. \end{equation*}

Real case. Assume H is real. Then f_A is symmetric. From the real polarization identity (cf.  Positive operators I, Eq.~(RSPI)) we have

(19)   \begin{equation*} 4f_A(x,y)=f_A(x+y,x+y)-f_A(x-y,x-y), \end{equation*}

and therefore

(20)   \begin{equation*} \vert f_A(x,y)\vert =\frac{1}{4}\vert f_A(x+y,x+y)-f_A(x-y,x-y)\vert. \end{equation*}

For any real a,b we have \vert a-b\vert\leq \vert a\vert+\vert b\vert. Therefore

(21)   \begin{equation*} \vert f_A(x,y)\vert \leq \frac{1}{4}\big(\vert f_A(x+y,x+y)\vert +\vert f_A(x-y,x-y)\vert\big). \end{equation*}

By (18), with z=x+y and z=x-y, we obtain

(22)   \begin{equation*} \vert f_A(x+y,x+y)\vert\leq m \Vert x+y\Vert^2,\qquad \vert f_A(x-y,x-y)\vert\leq m\Vert x-y\Vert^2. \end{equation*}

Thus

(23)   \begin{equation*} \vert f_A(x,y)\vert \leq \frac{1}{4}m\left(\Vert x+y\Vert^2+\Vert x-y\Vert^2\right). \end{equation*}

Using the Parallelogram Law we obtain

(24)   \begin{equation*} \vert(x,Ay)\vert=\vert f_A(x,y)\vert \leq \frac12 m\left(\Vert x\Vert^2+\Vert y\Vert^2\right). \end{equation*}

Now use the characterization of \Vert A\Vert via f_A:

(25)   \begin{equation*} \Vert A\Vert =\Vert f_A\Vert =\sup_{\Vert x\Vert=1,\Vert y\Vert=1}\vert(x,Ay)\vert \leq \frac{m}{2} \sup_{\Vert x\Vert=1,\Vert y\Vert=1} \left(\Vert x\Vert^2+\Vert y\Vert^2\right) = m. \end{equation*}

Recalling the definition (17) of m we obtain the desired opposite
inequality

(26)   \begin{equation*} \Vert A\Vert\leq \sup_{\Vert x\Vert=1}\vert (x,Ax)\vert. \end{equation*}

Complex case. Now let H be complex. We again want to estimate \vert (x,Ay)\vert.
Given any x\in H, we can always change its phase so that the scalar product (x,Ay) becomes real and non–negative. More precisely, if (x,Ay)\neq 0, choose

    \[ c = \frac{\overline{(x,Ay)}}{\vert (x,Ay)\vert}, \]

so that c has modulus 1 and (cx,Ay)\geq 0 is real. If (x,Ay)=0, there is nothing to prove. Let therefore x'=cx, with \vert c\vert=1, so that (x',Ay) is
non–negative. Then

(27)   \begin{equation*} \vert (x,Ay)\vert=\vert (x',Ay)\vert=(x',Ay). \end{equation*}

We now apply the complex polarization identity to (x',Ay). The right–hand side of this identity contains both real and imaginary parts. Since the left–hand side is real, the imaginary parts on the right–hand side must cancel, while the real parts have exactly the same form as in the previously considered real case. Therefore, repeating the preceding argument with x' in place of x, we obtain

(28)   \begin{equation*} \vert(x',Ay)\vert \leq \frac12 m\left(\Vert x'\Vert^2+\Vert y\Vert^2\right). \end{equation*}

But \Vert x'\Vert =\Vert x\Vert, and \vert(x',Ay)\vert=\vert(x,Ay)\vert.
Thus again

(29)   \begin{equation*} \Vert A\Vert=\Vert f_A\Vert \leq m=\sup_{\Vert x\Vert=1}\vert(x,Ax)\vert. \end{equation*}

Combining this with the first inequality finishes the proof.

Corollary 1. Let A be a bounded self–adjoint positive operator on a pre–Hilbert space
H. Then

(30)   \begin{equation*} \Vert A\Vert = \sup_{\Vert x\Vert=1}(x,Ax). \end{equation*}

Proof.  By Proposition 3 we know that

    \[ \Vert A\Vert =\sup_{\Vert x\Vert=1}\vert(x,Ax)\vert. \]

If A is positive, then (x,Ax)\geq 0 for all x, so \vert(x,Ax)\vert=(x,Ax) and we obtain the desired formula.

Corollary 2. Let A be a bounded self–adjoint operator on a pre–Hilbert space
H. Then

(31)   \begin{equation*} \Vert A^2\Vert = \Vert A\Vert^2. \end{equation*}

Proof. Using Proposition 3 ,we have

    \[ \Vert A^2\Vert = \sup_{\Vert x\Vert=1}\vert(x,A^2x)\vert= \sup_{\Vert x\Vert=1}\vert(Ax,Ax)\vert= \sup_{\Vert x\Vert=1}\Vert Ax\Vert^2=\Vert A\Vert^2.\]

=References.
[1] Bourbaki, N., Topological Vector Spaces, Chapters 1–5, Springer, 2003.

Afternotes.

12-06-26 Preparing the next note I was having a chat with Perplexity AI. Here is an excerpt:

I: “But you also wrote “while at x=1 the terms do not even tend to zero, so the series diverges there”
How would you explain this statement?”

Perplexity AI: That statement was simply wrong; there is no way to rescue it mathematically once we look carefully at the coefficients.

No shame whatsoever!

16-06-26 Removed Exercise 4, added Corollary 2.

Print 🖨 PDF 📄 eBook 📱
This entry was posted in Functional analysis, Hilbert space and tagged , , . Bookmark the permalink.

Leave a Reply