In linear algebra, a frame of a vector space V with an inner product can be seen as a generalization of the idea of a basis to sets which may be linearly dependent. The key issue related to the construction of a frame appears when we have a sequence of vectors , with each and we want to express an arbitrary element as a linear combination of the vectors :
and want to determine the coefficients . If the set does not span , then these coefficients cannot be determined for all such . If spans and also is linearly independent, this set forms a basis of , and the coefficients are uniquely determined by : they are the coordinates of relative to this basis. If, however, spans but is not linearly independent, the question of how to determine the coefficients becomes less apparent, in particular if is of infinite dimension.
Given that spans and is linearly dependent, it may appear obvious that we should remove vectors from the set until it becomes linearly independent and forms a basis. There are some problems with this strategy:
In 1952, Duffin and Schaeffer gave a solution to this problem, by describing a condition on the set that makes it possible to compute the coefficients in a simple way. More precisely, a frame is a set of elements of V which satisfy the so-called frame condition:
The numbers A and B are called lower and upper frame bounds.
It can be shown that the frame condition is sufficient to be able to find a set of dual frame vectors with the property that
for any . This implies that a frame together with its dual frame has the same property as a basis and its dual basis in terms of reconstructing a vector from scalar products.
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If the set is a frame of V, it spans V. Otherwise there would exist at least one non-zero which would be orthogonal to all . If we insert into the frame condition, we obtain
therefore , which is a violation of the initial assumptions on the lower frame bound.
If a set of vectors spans V, this is not a sufficient condition for calling the set a frame. As an example, consider and the infinite set given by
This set spans V but since we cannot choose . Consequently, the set is not a frame.
A frame is tight if the frame bounds and are equal. This means that the frame obeys a generalized Parseval's identity. For example, the union of orthonormal bases of a vector space forms a tight frame with . If , then a frame is either called normalized or Parseval. However, some of the literature refers to a frame for which where is a constant independent of (see uniform below) as a normalized frame.
A frame is uniform if each element has the same norm: where is a constant independent of . A uniform normalized tight frame with is an orthonormal basis.
The frame condition is both sufficient and necessary for allowing the construction of a dual or conjugate frame, , relative the original frame, . The duality of this frame implies that
is satisfied for all . In order to construct the dual frame, we first need the linear mapping: defined as
From this definition of and linearity in the first argument of the inner product, it now follows that
which can be inserted into the frame condition to get
The properties of can be summarised as follows:
The dual frame is defined by mapping each element of the frame with :
To see that this make sense, let be arbitrary and set
It is then the case that
which proves that
Alternatively, we can set
By inserting the above definition of and applying known properties of and its inverse, we get
which shows that
This derivation of the dual frame is a summary of section 3 in the article by Duffin and Schaeffer. They use the term conjugate frame for what here is called dual frame.
Frames were introduced by Duffin and Schaeffer in their study on nonharmonic Fourier series. They remained obscure until Mallat, Daubechies, and others used them to analyze wavelets in the 1980s. Some practical uses of frames today include robust coding and design and analysis of filter banks.