# Fréchet derivative is unique

Theorem The Fréchet derivative is unique.
Proof. Assume that both $A$ and $B$ in $L(\mathsf{V,W})$ satisfy the condition for the Fréchet derivative (http://planetmath.org/derivative2) at the point $\mathbf{x}$. To prove that they are equal we will show that for all $\varepsilon>0$ the operator norm  $\|A-B\|$ is not greater than $\varepsilon$. By the definition of limit there exists a positive  $\delta$ such that for all $\|\mathbf{h}\|\leq\delta$

 $\|f(\mathbf{x}+\mathbf{h})-f(\mathbf{x})-A\mathbf{h}\|\leq\frac{\varepsilon}{2% }\cdot\|\mathbf{h}\|\mbox{ and }\|f(\mathbf{x}+\mathbf{h})-f(\mathbf{x})-B% \mathbf{h}\|\leq\frac{\varepsilon}{2}\cdot\|\mathbf{h}\|$

holds. This gives

 $\displaystyle\|(A-B)\mathbf{h}\|$ $\displaystyle=\|(f(\mathbf{x}+\mathbf{h})-f(\mathbf{x})-A\mathbf{h})-(f(% \mathbf{x}+\mathbf{h})-f(\mathbf{x})-B\mathbf{h})\|$ $\displaystyle\leq\|f(\mathbf{x}+\mathbf{h})-f(\mathbf{x})-A\mathbf{h}\|+\|f(% \mathbf{x}+\mathbf{h})-f(\mathbf{x})-B\mathbf{h}\|$ $\displaystyle<\varepsilon\cdot\|\mathbf{h}\|.$

Now we have

 $\delta\cdot\|A-B\|=\delta\cdot\sup_{\|\mathbf{g}\|\leq 1}\|(A-B)\mathbf{g}\|=% \sup_{\|\mathbf{g}\|\leq\delta}\|(A-B)\mathbf{g}\|\leq\sup_{\|\mathbf{g}\|\leq% \delta}\varepsilon\cdot\|\mathbf{g}\|\leq\varepsilon\cdot\delta,$

thus $\|A-B\|\leq\varepsilon$ as we wanted to show.

Title Fréchet derivative is unique FrechetDerivativeIsUnique 2013-03-22 16:08:35 2013-03-22 16:08:35 Mathprof (13753) Mathprof (13753) 12 Mathprof (13753) Theorem msc 46G05 derivative