Properties of series $\sum_{k=0}^n A^k(h_{n-k})$

Question

Let $A$ be an operator from a Hilbert space $H$ into itself, and denote $A^k$ the composition of $A$ with itself $k$ times as usual. Suppose $h_1, ..., h_n \in H$.

I am wondering if there's any properties for the series (or alternate expressions)

$$\sum_{k=0}^n A^k(h_{n-k}) = h_n + Ah_{n-1} + ... A^nh_0$$

or if anyone knows what it's called. I want to study it for my particular $A$ so I thought someone may have looked at such series before. In particular I am interested in the well-definedness of the limit $n \to \infty$ of the above expression.

Answer 1

If you want it to be well-defined, you should have the sum to be convergent to an element of $H$. As Hilbert space is a special case of Banach space, you can define a norm $||.||$ for elements of $H$, according to the inner product. Now, you may know that there is a theorem that says: "A partial sum is convergent, if it is absolutely convergent".

Therefore, you may show that

$\sum_{k=0}^n ||A^k(h_{n-k})|| = ||h_n|| + ||Ah_{n-1}|| + ... +||A^nh_0||$

which is a partial sum of real numbers, is convergent to a number.

I suppose, we need to have a special operator $A$ to make sure the whole thing works. We need $||A^n|| \rightarrow 0$, as $n\rightarrow \infty$. Then, having the inequality

$||A^nh_0|| \leq ||A^n||.||h_0||$

We can force $||A^nh_0||$ to go to zero.

Note that it is a necessary condition and it is not sufficient. To be more accurate, $||A^nh_0||\rightarrow 0$ is necessary. But, in order to have the sum to go to zero, we need to take more measurements.

Now, the question is: "How to find such an operator $A$?"

For finite dimensional spaces, like simple vector spaces, you may find matrices, with all eigenvalues having an absolute-value less than $1$. It guarantees the norm to be decreased, as $n$ increases.

For infinite dimensional cases, you can adapt a similar approach. I suppose, it gets a bit complicated. However, I do not have much information.

Hope it helps.