(Frame property 1) If
satisfies the property (
Frame formula 1
) then
for
where
Proof
We
calculate
hence
Make change
and use
.
and we arrive to the desired conclusion.
Proposition
(Frame property 2)
1. If a function
satisfies the formula (
Frame formula 1
) then
there exist
such that
we
have
(Riesz property)
2. If a function
has compact support and
satisfies the formula (
Riesz property
) then it
satisfies the formula (
Frame formula 1
).
Proof
(1). For a function
we
calculate
We use the formula (
Property of
scale and transport
4
).
We continue calculation of
We make a change
.
The function
has period 1.
The function
has period 1 and the above expression is the Fourier coefficient of it. By
proposition (
Parseval equality
) we
have
We substitute the formula (
Frame formula
1
):
so
that
and use the following consequence of the proposition
(
Frame property
1
):
thus
Proof
(2) According to the proposition (
Property
of transport 2
), the
function
takes the
form
for some finite sequence of real numbers
.
For this reason we already
have
To prove the
part
it suffices to disprove existence of
such that
.
This is so because
is periodic.
For
to vanish, every term
has to vanish. In other words, we aim to show that there cannot exist a
such
that
or
In the part (1) of the proof we obtained the generic relationship
.
We will reach our goal if we can find a sequence
:
where the
is Fourier transform of
:
Indeed, if such
exists then, according to
,
and we would have a contradiction with the formula
(
Riesz property
).
Existence of such
is standard. We approximate
on
by an
-indexed
sequence of periodic functions, obtain its Fourier coefficients
and use them in the combination
.
Proposition
(Frame property 3) If functions
satisfy the conditions
(a)
is compactly supported,
(b)
is a Riesz basis for
,
(c)
is biorthogonal to
.
then
(1)
(2) there exist
such that
Proof
It suffices to prove (1) and (2) for
.
Extending to the closure is a standard exercise because all involved
operations are
-continuous.