The real function space is a vector space because it is linear in multiplication and addition.
\(g(x)=cf(x)\)
\(h(x)=f(x)+k(x)\)
We start with our vector \(f(x)\).
\(h(x)=f(x)g(x)\)
The equivalent of the identity matrix is where \(g(x)=1\).
These are similar to endomorphisms where all off diagonal elements are \(0\).
\(h(x)=\dfrac{\delta }{\delta x}f(x)\)
\(h(x)=\int_{-\infty }^x f(z) dz\)
For a function \(v\) we can define operators \(Ov\).
Here we consider some examples and their properties.
\(Rv = rf(x)\)
This operator is hermitian. This is equivalent to a finite operator of the form \(rI\).
\(Xv = xf(x)\)
This operator is hermitian. This is equivalent to a finite operator of the form \(M_{ii}=i\) and \(M_{ij}=0\).
\(Dv = \dfrac{\delta }{\delta x}f(x)\)
While this operator is not hermitian, the following is:
\(-iDv = \dfrac{\delta }{\delta x}[-if(x)]\)
A form takes two vectors and produces a scalar.
We can use integration to get a bilinear form.
\(\int f(x) g(x) dx\)
If we instead want a sesquilinear form we can instead use:
\(\int \bar {f(x)} g(x) dx\)
Functionals map functions to scalars. They are the \(1\)-forms of infinite-dimensional vector spaces.
If we have a function \(f\), we can write functional \(J[f]\).
We can define neighbourhoods around a function \(f\). For example, taking \(y\) to be \(f\) with infintesimal changes. to each of the values.
The difference between the functional at both points is
\(\delta J=J[y]-J[f]\)
If
\(\delta J=J[y]-J[f]\)
is the same sign for all y around f, then J has an extremum at f.
A complete space with an inner product. That is, a Banach space where the norm is derived from an inner product.
Integrate over possible functions?
A complete normed vector space
For a vector in hermitian basis, for each eigenvector we have component. wave function is function on ith component.
The function is: \(\delta_{ij}\)
If \(i=j\) this is \(1\). Otherwise it is \(0\).
We introduced this in linear algebra.
The Dirac delta replaces the Kronecker delta for continuous functions.
That is, we want:
\(\delta (x\ne 0)=0\)
\(\delta (0)=+\infty\)
\(\int_{-\infty }^{\infty}\delta (x)dx=1\)