I realized that I have to diverge even more on my way to Krein spaces. We will need square roots of positive operators on real and complex Hilbert spaces, an so there is a need to discuss this subject in details. So here is the first part of this discussion. There will be more parts. How many more? I am not sure.
This note is based on the exposition of this subject in, chronologically, Refs. [1,2,3]. Ref. [4] follows the same idea, but using a different expansion. Most texts develop first the functional calculus, and consider the square root function as one of the whole 
algebra of continuous functions on the spectrum of a bounded operator. Here I am following closely Ref. [2] (Ref. [1] is using exactly the same method as [2]), as the reasoning there, although somewhat lengthy, is rather elementary.
Polarization identity
Let 
be a sesquilinear form on a vector space over the field 
.
Remark. In the real case “sesquilinear” is the same as bilinear. In the complex case I am using the convention in which a sesquilinear form is anti-linear in the first argument, and linear in the second argument.
Lemma 1. The following identity holds:
(1) 
(2) 
Proof. The proof is by expanding the right hand side using the properties of .
The form is called Hermitian if
(3) 
In the real case “Hermitian” is the same as “symmetric’‘. In that case the polarization identity reads
(RSPE) ![\[ 4f(x,y)=f(x+y,x+y)-f(x-y,x-y).\]](https://arkadiusz-jadczyk.eu/blog/wp-content/ql-cache/quicklatex.com-d397e8d0e329a67b0176a851fc4caf84_l3.png "Rendered by QuickLaTeX.com")
]Notice that, in the complex case, if is Hermitian, then the diagonal values 
are always real. Using the polarization identity we easily deduce that the converse is also true.
Proposition 1.
If 
and if is a sesquilinear form for which the diagonal values are real, then is Hermitian.
Proof. Taking into account the fact that the diagonal values of on the right hand side of (2) are real, we take the complex conjugate of Eq. (2) to obtain:
(4) 
This coincides with 
if we use yhe identities 
and 
, valid for any 
.
Positive operators
Pre-Hilbert space
First we will discuss a general pre-Hilbert case, and only later we will restrict ourselves to a Hilbert space, where stronger results can be obtained. Let 
be a pre-Hilbert space, real or complex. We denote by 
the algebra of all bounded linear operators on 
.
Definition 1. Let 
two self-adjoint bounded operators. We write 
(or 
) if, for all 
we have
(5) 
We call 
positive if 
.
The relation “
” is a partial order.
It follows immediately from this definition that the relation “” is a partial order. Moreover, if and 
are positive, and if 
then 
and 
are also positive.
In the complex case, for to be positive, we do not need to assume that is self–adjoint, as it follows already from the condition 
for all 
In fact we have the following
Proposition 2. Assume that is complex. With 
if 
is real for all 
, then is self–adjoint: 
Proof. If we set
(6) 
then 
is clearly a sesquilinear form We notice that is Hermitian if and only if is self–adjoint. Indeed, we have
(7) 
and 
. Thus the condition of Hermicity (3) for becomes
(8) 
and this is precisely the condition for to be self–adjoint.
Afternotes.
11-06-26 There was an error in my first version of the real polarization identity. Now corrected.
To be continued
References
[1][ Kreyshig, e., “Introductory Functional Analysis with Applications”, Wiley 1978.
[2] Lusternik, L.A., Sobolev, V.J., “Elements of Functional Analysis”, Wiley 1974.
[3] Petersen, G.K., “Analysis Now”, Springer 1989.
[4] Müger, M., “Introduction to Functional Analysis”, (2024)
https://www.math.ru.nl/~mueger/functionalanalysis.pdf.
