= Dual norm as a worst-case linear error bound
{title2=$|a^T\delta x|\le\|a\|_*\varepsilon$}
Suppose a perturbation $\delta x$ satisfies $\|\delta x\|\le\varepsilon$, with $\varepsilon\ge0$. <Absolute homogeneity of a norm> and the <dual norm> definition give
$$
|p(x+\delta x)-p(x)|=|p(\delta x)|
\le\|p\|_*\|\delta x\|\le\varepsilon\|p\|_*.
$$
For nonzero $\delta x$, divide it by its <norm> to obtain an input in the <unit ball>; for zero perturbation the inequality is immediate. This is an exact <error bound> for a <linear functional>, for every perturbation in the stated input set.
For $p(x)=2x_1-x_2$ and a budget $\varepsilon=0.1$, the sharp output-error bounds are \b[$0.2$] for $\|\delta x\|_1\le0.1$, \b[$\sqrt5/10\simeq0.224$] for $\|\delta x\|_2\le0.1$, and \b[$0.3$] for $\|\delta x\|_\infty\le0.1$. Scale the maximizing <vectors> from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The <L1 norm> budget constrains a total absolute error; the <maximum norm> budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.
\Image[/dual-norm-sharp-linear-error-budgets.png]
{title=<Dual norms> give exact worst-case errors for the same <linear functional>}
{description=Original bounds for $p(x)=2x_1-x_2$. Lines show the sharp error magnitude as the stated input <norm> budget increases. The right panel gives the full possible error intervals at $\varepsilon=0.1$; these are the full feasible intervals for the stated deterministic bounds.}
The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear <differentiable function> with scalar values, its <Fréchet derivative> at the base point is a <linear functional> governing the first-order response. An exact finite-perturbation <error bound> additionally needs control of the <Taylor remainder> or of the <Fréchet derivative> along the path.
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