EDIT: user14717's comments resolved the conflict described below. The choice of the second norm significantly limits set of perturbations which are considered. The set is reduced from those perturbations which are small, to those perturbations which are small and also have a small derivative. There is no longer any conflict and differentiation remains ill-conditioned when very general perturbations are allowed.
ORIGINAL:
Is the problem of computing the derivative of a function an ill-conditioned problem?
I believe that answer is yes, but I am not certain because it is possible to choose the topology such that the answer is no.
I choose to view differentiation as the function $D$ which maps $f \in U = C^1([a,b],\mathbb{R})$ into $f' \in V = C([a,b],\mathbb{R})$. Now the conditioning of a function should reflect the sensitivity of the output to small changes in the input. I therefore choose to consider the auxiliary function \begin{equation} \kappa_D(f,\delta) = \sup \left\{ \frac{ \| f' - g'\|_V }{\| f - g\|_U} \: : \: 0 < \|f-g\|_U < \delta \right \}. \end{equation} It has the nice property that $\delta \rightarrow \kappa_D(f,\delta)$ is nonnegative and nondecreasing, hence the limit \begin{equation} \underset{\delta \rightarrow 0_+}{\lim} \kappa_D(f,\delta). \end{equation} exists and I can define the condition number of $D$ at the point $f$ by \begin{equation} \kappa_D(f) = \underset{\delta \rightarrow 0_+}{\lim} \kappa_D(f,\delta) \end{equation} This construction mirrors the construction used to define the absolute condition number of a function $f : \mathbb{R} \rightarrow \mathbb{R}$ at a point $x \in \mathbb{R}$. Now what is the actual value of $\kappa_D(f)$? That depends very much on the norms used inside $U$ and $V$. My first choice is to treat $U$ as a subspace of $V$ and use the infinity norm. Given $f \in U$ and any $\delta > 0$, I consider $g$ such that \begin{equation} f(x) - g(x) = \rho \sin \left( \frac{x}{\rho} \right) \end{equation} where I choose $\rho \in (0,\delta)$. Then $\|f-g\|_\infty = \rho$ but $\|f'-g'\|_\infty = 1$. It follows that \begin{equation} \kappa_D(f,\delta) \ge \rho^{-1} \end{equation} and since $\rho \in (0,\delta)$ was arbitrary we have \begin{equation} \kappa_D(f,\delta) = \infty. \end{equation} Obviously, we also have \begin{equation} \kappa_D(f) = \infty. \end{equation} However, this conclusion hinges on my choice of norm. If I instead define \begin{equation} \| f - g\|_U = \|f-g\|_\infty + \|f'-g'\|_\infty \end{equation} and maintain \begin{equation} \| f' - g'\|_V = \|f' - g'\|_\infty \end{equation} then I always have \begin{equation} \frac{\|f'-g'\|_V}{\|f - g\|_U} = \frac{\|f'-g'\|_\infty}{\|f-g\|_\infty + \|f'-g'\|_\infty} \leq 1, \end{equation} from which it immediately follows that \begin{equation} \kappa_D(f) \leq 1. \end{equation} What would I like to see in answer? I would like to understand if differentiation is ill-conditioned or not. I have made many choices along the way. I would like to understand if they are reasonable or not?