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I am wondering if there is a theorem in matrix theory as an analog to Banach-Steinhaus theorem. Here are some attempts. Suppose $D_n\in\mathbb{R^{d\times d}}$, $n=1,2,\ldots$ is a sequence of matrix and for every $x\in\mathbb{R}^d$, there exists some constant $c_x>0$, such that $\|D_n x\|\leq c_x$. From Banach-Steinhaus theorem, there exists a constant $C>0$ such that $\|D_n\|\leq C$.

  1. Is the condition “$\sup\limits_n\|D_n x\|<\infty$” equivalent to “the eigenvalues of $D_n$ has a uniform bound”?
  2. For finite dimensional case, is there any simple proof for Banach-Steinhaus theorem?

Thanks.

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First of all, the domain and codomain of the maps in the Banach-Steinhaus theorem do not have to be the same. So if you want to translate it into matrices, there's no reason to restrict to square matrices. This also means it is not very natural to try to formulate things in terms of eigenvalues.

In any case, having a uniform bound on the eigenvalues of the $D_n$ is weaker than being pointwise bounded. For instance, consider the $2\times 2$ matrices $$D_n=\begin{pmatrix}1 & n \\ 0 & 0\end{pmatrix}.$$ These matrices are all projections so their eigenvalues are only $0$ and $1$ but $D_nx$ is unbounded for $x=(0,1)$.

In finite dimensions, the Banach-Steinhaus theorem is essentially trivial. For instance, a uniform bound on $\|D_n e_i\|$ where $e_i$ is the $i$th standard basis vector gives a uniform bound on the entries of the $i$th column of $D_n$. Taking this for $i=1,\dots,d$, you get a uniform bound on all the entries of $D_n$ and hence on its norm (using whatever norm you want since all norms are equivalent in finite dimensions).

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