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Is there any meaningful sense in which the backshift operator can be said to have a derivative?

Here's my attempt at constructing one. Based on some helpful threads on this site, I represent the backshift operator $B$ as a matrix (showed here for an example time series of length $3$): $$ \begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix} = \begin{pmatrix} 0\\ y_1\\ y_2 \end{pmatrix}\ . $$ I can take the derivative of the above equation with respect to $y_1$ and $y_2$ to obtain the vectors $(0, 1, 0)$ and $(0, 0, 1)$ respectively (the derivative w.r.t $y_3$ is undefined.) Extrapolating, and wishing to obtain an operator (instead of a vector) from the differentiation process, I conclude that the derivative of the backshift operator is a type of "shifted" indicator function $I_{t}$ which has a straightforward matrix representation. For example, I would write $$ \frac{\partial B}{\partial y_1} = I_1 = \begin{pmatrix} 0 & 0 & 0\\ 1/y_1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\ . $$ Does this calculation have any meaning, or are there some operator-space/common-sense rules which forbid some steps? References would be appreciated.

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    $\begingroup$ The derivative with respect to $y_3$ is certainly defined: it's the zero vector. $\endgroup$ Commented Oct 20, 2022 at 6:30
  • $\begingroup$ @GregMartin Thanks for the correction. $\endgroup$
    – Anthony
    Commented Oct 20, 2022 at 8:53

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The backshift operator is a mapping (an "operator") between vector spaces, namely spaces of time series or sequences,

$$ B\colon \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}, (y_i)\mapsto (y_{i-1}). $$

Here, $\mathbb{R}^\mathbb{N}$ is the space of mappings from $\mathbb{N}$ to $\mathbb{R}$, i.e., of real-valued sequences indexed by natural numbers. Time series that are not infinite can be accommodated by having only finitely many observations nonzero. The space $\mathbb{R}^\mathbb{N}$ is naturally a real vector space: we can add time series and multiply them by real scalars.

In functional analysis, there is a notion of differential of an operator between normed vector spaces. If an operator $f$ admits a "local linear approximation" $\varphi$ near a point $x$, then we say that $f$ is differentiable and that $\varphi$ is its differential at $x$ (see, e.g., Coleman, 2012, section 2.2). Note that this generalizes the familiar notion of differentiability of mappings between finite dimensional spaces: a function $f\colon\mathbb{R}^n\to\mathbb{R}^m$ is differentiable at a point $x$ if and only if it admits a well-defined tangential subspace (tangent line in the most familiar case of $n=m=1$) near $x$.

Now, we observe that $B$ is already linear:

$$ B\big(\lambda(y_i)\big) = \lambda B\big((y_i)\big)\quad\text{and}\quad B\big((y_i)+(z_i)\big) = B\big((y_i)\big)+B\big((z_i)\big). $$

Thus, a (unique!) "best approximation" to $B$ is $B$ itself (see here at Math.SE). Therefore, the differential of $B$ at all "points" (i.e., sequences) is $B$ itself. Note that we don't even need to think about what norm we put on our space $\mathbb{R}^\mathbb{N}$.

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