Wikipedia lists the condition number at x of any function $f: V_1 \rightarrow V_2$ between Banach spaces to be $\lim_{\epsilon\rightarrow 0^+} \sup_{||\delta x|| \le \epsilon} \left[ \frac{||f(x+\delta x) -f(x)||}{||f(x)||} / \frac{||\delta x||}{||x||}\right]$
However, it seems to me that this would give any linear operator a condition number of one? So I'm having trouble reconciling this with the notion of matrix condition number
If $$ f(x):=A^{-1}x, $$ then $$ \frac{\|f(x+\delta x)-f(x)\|}{\|f(x)\|}/ \frac{\|\delta x\|}{\|x\|} = \frac{\|A^{-1}(x+\delta x)-A^{-1}x\|}{\|A^{-1}x\|}/ \frac{\|\delta x\|}{\|x\|} = \frac{\|A^{-1}\delta x\|}{\|\delta x\|} \frac{\|x\|}{\|A^{-1}x\|}, $$ so $$ \lim_{\epsilon\rightarrow 0^+}\sup_{\|\delta x\|\leq\epsilon} \frac{\|A^{-1}\delta x\|}{\|\delta x\|} \frac{\|x\|}{\|A^{-1}x\|} = \|A^{-1}\| \frac{\|x\|}{\|A^{-1}x\|}. $$ Note that the condition number depends on the "right-hand side" $x$. To eliminate this dependency and thus to obtain the "worst-case" condition number, we want to maximize it over all nonzero $x$ which gives the classical matrix condition number: $$ \sup_{x\neq 0} \|A^{-1}\| \frac{\|x\|}{\|A^{-1}x\|} = \|A^{-1}\| \sup_{y\neq 0} \frac{\|Ay\|}{\|y\|} =\|A\|\|A^{-1}\|. $$
