I'm trying to grasp some understanding of the concept of stability in the context of the frame theory. The definition of frame in ref1 states that a sequence of functions $\{ \phi_n \}_{n \in \Gamma}$ belonging to a dictionary $D$ is a frame if there exist two constants $B \geq A > 0 $ so that:

\begin{equation}\tag{1} A \|f\|^2 \leq \sum_{n\in\Gamma} |\langle f, \phi_n \rangle|^2 \leq B \|f\|^2 \end{equation}

where $f$ belongs to a Hilbert space $H$.

From my understanding, the definition of frame entails stability since the variation of the norm of the representation $ \|\Phi(f)\|^2 = \sum_{\Gamma} |\langle f, \phi_n \rangle |^2$ is tied to the variation of $\|f\|^2$ and bounded by the constants $A$ and $B$. In ref2, the concept of stability is explained with reference to the time-warped deformations $f_\tau(t) = f(t - \tau(t)) $ saying that, to be stable, there should be a constant $C>0$ so that:

\begin{equation}\tag{2} \| \Phi(f) - \Phi(f_\tau) \| \leq C \sup_t |\tau'(t)| \|f\| \end{equation}

It is also shown that, in the case of the Fourier transform, $\| \Phi(f) - \Phi(f_\tau) \|$ do not to decrease proportionally to $\epsilon$ ($\tau(t) = \epsilon t$) and equation (2) cannot be satisfied.

The first question is: to what extent conditions (1) and (2) are equivalent?

The second question relates to the STFT. In 1, section 4.2.1 says that the STFT is stable. Referring to the results in 2 mentioned above, my understanding is that the stability comes from the fact that the window operates some kind of averaging over a finite interval of frequencies. However, for windows very long in time, I was expecting to find again the same kind of instability seen for the Fourier transform. What am I missing here?


1 Answer 1


They aren't equivalent; "stability" is used differently in each context.

$(1)$ guarantees a stable inverse. If $A=0$, we lose information. Existence of such $A$ and $B$ ensure the representation's norm is bound by input's norm alone, and not its other characteristics (e.g. regularity, frequency).

$(2)$ guarantees a stable representation, in that the distance between representations of deformed and undeformed inputs grows by, and only by, the size of deformation. STFT is stable per $(1)$ but not per $(2)$.

In context of scattering, $(1)$ is necessary to guarantee $(2)$, and we must also have $B \leq 1$, else cascaded operators explode the input norm. Explained further here and here.

  • 1
    $\begingroup$ You beat me to it. Very nice answer :-). $\endgroup$
    – Royi
    Dec 12, 2021 at 10:53
  • $\begingroup$ Thanks! Clear explanation :) $\endgroup$
    – dac
    Dec 14, 2021 at 2:10

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