I'm trying to determine an upper bound for the relative error when solving a linear system with a perturbed input. Particularly, I consider a linear system of the form: \begin{equation}Ax = b\end{equation} In presence of a perturbation $\delta b$ of the input term $b$, the linear system solution is altered as: \begin{equation}A(x+\delta x) = b+\delta b\end{equation}
The typical (and best to my knowledge) bound that can be obtained in absence of other information is: \begin{equation} \frac{\|\delta x\|_2}{\|x\|_2} \leq \frac{\sigma_1}{\sigma_n}\frac{\|\delta b\|_2}{\|b\|_2} = \kappa \frac{\|\delta b\|_2}{\|b\|_2}\end{equation} where $\sigma_1, \sigma_n$ are the maximal and minimal singular values of $A$, and thus $\kappa$ is the 2-norm condition number of $A$. The bound is sharp and verified when: \begin{equation}b = \alpha u_1\end{equation} \begin{equation}\delta b = \beta u_n\end{equation} where $u_1, u_n$ are the first and last columns of the (ordered) $U$ matrix of the SVD decomposition of $A$, \begin{equation} A = U\Sigma V^T\end{equation}
In my case, $b$ and $\delta b$ have peculiar forms, as they both can be expressed as the rescaled version of a primitive vector $b'$, $\delta b'$ with respect to their maximum absolute value, i.e.: \begin{equation} b = \frac{b'}{\|b'\|_\infty}c\end{equation} \begin{equation} \delta b = \frac{\delta b'}{\|\delta b'\|_\infty}\delta c\end{equation}
Plugging in their definitions in the previous bound returns: \begin{equation} \frac{\|\delta x\|_2}{\| x\|_2} \leq \kappa \frac{\|\delta b'\|_2}{\|b'\|_2} \frac{\|b'\|_\infty}{\|\delta b'\|_\infty} \frac{\delta c}{c}\end{equation}
My questions now are: (1) is this the best (i.e., more eloquent yet sharp) form of the bound that can be found? (2) is the previous worst-case configuration still applying?
I tried two different strategies.
The first was to keep on majorizing by considering that: \begin{equation} \|\delta b'\|_2 \leq \sqrt{n} \|\delta b'\|_\infty \end{equation} \begin{equation} \|b'\|_2 \geq \|b'\|_\infty \end{equation} and thus: \begin{equation} \frac{\|\delta x\|_2}{\| x\|_2} \leq \kappa \sqrt{n} \frac{\delta c}{c}\end{equation}
However, I fear that this bound may now not be sharp anymore, and I can't intuitively assess which would be the updated worst-case configuration realizing the bound.
Alternatively, since the previous worst-case conditions was derived under general assumptions, it should still hold now. In that case, plugging in the definitions of $b$ and $\delta b$ allows to evaluate the two coefficients $\alpha$, $\beta$: \begin{equation}\frac{b'}{\|b'\|_\infty} = \frac{b}{c} = \frac{\alpha u_1}{c}\end{equation} Since the left-hand side vector has $1$ as its maximum element, the same must hold for the right hand-side, and thus: \begin{equation} \alpha = \frac{c}{\| u_1\|_\infty}\end{equation} With a similar reasoning: \begin{equation} \beta = \frac{\delta c}{\| u_n\|_\infty}\end{equation}
And thus the bound becomes: \begin{equation} \frac{\|\delta x\|_2}{\| x\|_2} \leq \kappa \frac{\| u_1 \|_\infty}{\| u_n \|_\infty}\frac{\delta c}{c}\end{equation}
which seems to be sharper as the dependence on $\sqrt{n}$ is lost, but I'm not sure whether it was valid in the beginning to start assuming the previous worst-case still holds.