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The normalized Hermitian form on a finite function space

Statement

Let X be a nonempty finite set. On CX, with pointwise vector-space operations, define f,h=X1xXf(x)h(x). The denominator is the positive real image of X. This is an inner product, linear in the first variable.

Facts & Assumptions

Given: X finite and nonempty, m=X>0 viewed in RC, and functions f,h:XC.

[F1]

The linear-first inner-product axioms are linearity in the first variable, conjugate symmetry, and positive definiteness (Real and complex inner product spaces, with the inner product linear in the first argument).

[F2]

For z=a+bi, z=abi and z=a2+b2 (Real and imaginary parts, complex conjugation, and modulus).

[F3]

Finite real sums are defined by enumeration, independently of that enumeration (The sum iSai over a finite index set, and its product form).

[F4]

A finite sum of nonnegative reals is nonnegative and is zero only if every summand is zero (Laws of finite sums and finite products).

[F5]

Finite monoid sums are independent of enumeration and agree with real sums on real summands (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[F6]

Conjugation preserves addition and multiplication, is involutive, and zz=z20 with equality exactly at z=0 (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

[F7]

A property holding at zero and preserved under successor holds for every natural number (The principle of mathematical induction).

Proof

technique · direct
1.1

Pointwise addition and scaling make CX a vector space: each abelian addition identity, both distributive identities, scalar associativity and the scalar identity hold at each x by the field laws in C. F5 defines every complex sum in the displayed form. Since m>0, m1 exists and is positive real, and F2 gives m1=m1. Thus the form is defined on every pair of functions.

F5F2givenalgebra
1.2

For complex lists aj,bj and cC, finite sums satisfy j(aj+bj)=jaj+jbj, jcaj=cjaj and jaj=jaj. Here is the induction verifying their complex types: at length zero all sums vanish and 0=0. On appending a,b, the first formula follows by rearranging (A+B)+(a+b)=(A+a)+(B+b); the second from cA+ca=c(A+a); the third from A+a=A+a. F7 proves the three identities for every length, and F5 transfers them to any finite enumeration of X.

F5F6F7algebra
2.1

For a,bC and f1,f2,hCX, expand the summand and apply step 1.2 to obtain af1+bf2,h=m1x(af1(x)+bf2(x))h(x)=af1,h+bf2,h.

step 1.2algebra
2.2

Applying conjugation to the finite sum and using its involution gives h,f=m1xh(x)f(x)=f,h.

step 1.1step 1.2F6algebra
2.3

On the diagonal, F6 gives f,f=m1xf(x)2. These are nonnegative real summands, so F5 identifies their sum with the real finite sum of F3, and F4 shows the result is real and nonnegative. If it is zero, multiplication by m>0 gives xf(x)2=0; F4 forces each f(x)2=0, hence f(x)=0 by F6. Conversely f=0 makes every summand zero.

F6F5F3F4step 1.1
3.1

Steps 2.1–2.3 verify F1 and therefore give an inner product. If X={x} the expression is f(x)h(x) and the same verification applies. The zero function has diagonal value zero by step 2.3; the empty set is excluded precisely because the prescribed normalization would divide by zero.

F1step 2.1step 2.2step 2.3

Sources

Axler, 6.2–6.3(a),(b), pp. 183–184, fixes the linear-first convention and positive weights. Etingof et al., §4.5 opening, p. 67, is the class-function specialization. The complex finite-sum laws and definiteness are explicitly derived above.

Depends on

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Sources