(Frame property 1) Assume that
satisfies the property (
Frame formula 1
). Then
for
we
have
where
Proof
We
have
hence
Make change
and use
.
and we arrive to the desired conclusion.
Proposition
(Frame property 2)
1. Assume that function
satisfies the formula (
Frame formula 1
) then
there exist
such that
we
have
(Riesz property)
2. Assume that function
has compact support and satisfies the formula
(
Riesz property
) then it satisfies the formula
(
Frame formula 1
).
Proof
(1). For an
we
calculate
We use the formula (
Property of
scale and transport
4
).
We continue calculation of
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
Suppose the contrary, there exists a
with the property
.
Let
be a large parameter. We subsitute
into the formula (
Riesz
property
):
We
have
As we put
the LHS of
blows up. On the RHS we have a sum of differentiable functions of
vanishing as
(by
)
and the number of terms in the sum grows linearly as
(by compactness of support of
).
Hence, the RHS of
is bounded or smaller as
.
We obtained a contradiction.
Proposition
(Frame property 3) Assume that functions
are such that
(a)
is compactly supported,
(b)
is a Riesz basis for
,
(c)
is biorthogonal to
.
Then
(1)
(2) there exist
such that
Proof
It is enough to prove (1) and (2) for
.
Extending to the closure is a standard exercise because all involved
operations are
-continuous.
