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$\DeclareMathOperator{\op}{\mathrm{op}}$Since matrix multiplication is continuous, I expect that if $A_n\to A\in \mathrm{Mat}_{d\times d}(\mathbb{R})$ for the operator-norm and if $x_n\to x$ in $\mathbb{R}^d$ then $$ \left\|A_n x_n - Ax\right\|\to 0. $$

However, I cannot manage to derive a reasonable bound for this, where by "reasonable" I mean that: $$ \left\| B y - Ax \right\| \leq p(\|B-A\|_{\op},\|y-x\|), $$ for some non-negative continuous function $p:\mathbb{R}^2\rightarrow [0,\infty)$ converging to $0$ as its arguments converge to $0$ and does not depend on the quantities $\|x\|$ and $\|y\|$.

To date, the best I have gotten is $$ \left\| B y - Ax \right\| \leq \min\left\{ \|B\|_{\op}\|x-y\| + \|A-B\|_{\op}\|y\| , \|A\|_{\op}\|x-y\| + \|A-B\|_{\op}\|x\| \right\}, $$ but this is of no use to me since the bound depends on $\|x\|$ and on $\|y\|$.

Are such bounds possible? What if I also assume that $A,B \in \mathrm{SO}(d,\mathbb{R})$?

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    $\begingroup$ Consider the case $x=y$. Then $By-Ax=(B-A)x$ so $\|By-Ax\|\le \|(B-A)\| \, \|x\|$ with equality if $x$ is an eigenvector of $B-A$ no matter how small $B-A$ is. So there is no bound like you suggested, but as Geoff said you don't need it just to prove convergence. $\endgroup$
    – Brendan McKay
    Commented Oct 6, 2020 at 10:43
  • $\begingroup$ You don't need what you ask for just to prove that $A_{n}x_{n} \to Ax$ when $A_{n} \to A$ and $x_{n} \to x$. Just write $A_{n}x_{n}- Ax$ as $(A_{n}-A)(x_{n}-x) + A(x_{n}-x) + (A_{n}-A)x.$ $\endgroup$
    – Geoff Robinson
    Commented Oct 6, 2020 at 10:51
  • $\begingroup$ Yes, to prove convergence I can just use the fact that function evaluation is continuous in the compact-open topology on locally-compact domains. But I was asking more to helop me derive some convergence bounds... $\endgroup$
    – ABIM
    Commented Oct 6, 2020 at 11:05

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