From the formula
we
conclude
We would like to take inverse Fourier transform of
.

By condition 1 and according to the section
(
Fourier transform of
delta function
),
For a general function
,
thus
We now take the inverse Fourier transform of the equality
.
The product becomes
convolution:
Thus we have the condition
(
Multiresolution
analysis
)-1:

We now verify the condition
(
Multiresolution analysis
)-3. It
suffices to show that
we
have
First, we consider function
with compact
support:
We
estimate
Thus

The set of functions
with compact support constitutes a dense set in
and for each function on such dense set we
have
Hence, the above convergence result extends to all
by the following standard argument. Let
and
have compact support and
,
as
.
If
does not converge to zero then there is a subsequence
such that
is separated from zero. But then we arrive to contradiction because everything
on the RHS of
can be arbitrarily small.