Consider the following quantity: $$ f_O(|\psi\rangle) := \langle \psi | O | \psi \rangle $$
How would we study a perturbation on $|\psi\rangle$, given that it has to be a valid quantum state? What mathematical tools underlie this quantity?
For example, we might be interested in understanding whether there are specific $|\psi\rangle$ with high sensitivity to disruption.
If you care more about upper bounds to $f_O$ under perturbations (i.e. how much a given perturbation can change $f_O$ in the worst case) then one can prove the following:
Result. For all observables $O$ and all pure states $|\psi\rangle,|\psi'\rangle$ it holds that $$ \big|f_O(|\psi\rangle)-f_O(|\psi'\rangle)\big|\leq 2\|O\|_\infty \sqrt{1-|\langle \psi|\psi'\rangle|^2}\tag1 $$ where $\|\cdot\|_\infty$ is the operator norm (largest singular value).
Perhaps counterintuivitely, this result is proven most easily by changing the language from pure states to general mixed states $\rho$, at least for a moment. First, we re-define $f_O(\rho):={\rm tr}(\rho O)$ and we note that this reproduces the original $f_O$ for the pure-state case $\rho=|\psi\rangle\langle\psi|$. The reason this is useful is the following standard argument: Given any two density matrices $\rho,\rho'$, applying a basic trace-norm inequality yields $$ |f_O(\rho)-f_O(\rho')|=|{\rm tr}((\rho-\rho')O)|\leq\|\rho-\rho'\|_1\|O\|_\infty\,. $$ (Here $\|\cdot\|_1$ is the trace norm = sum of all singular values).
Now we can substitute pure states for $\rho,\rho'$ which simplifies to the point where the trace norm disappears entirely. More precisely, given any two pure states $\psi,\psi'$ it holds that $$\|\,|\psi\rangle\langle\psi|-|\psi'\rangle\langle\psi'|\,\|_1=2\sqrt{1-|\langle \psi|\psi'\rangle|^2}$$ (cf. Eq. (1.186) in the book of Watrous / alt link) so combining these two identities yields Eq. (1), as desired.