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[parent] Fréchet derivative is unique (Theorem)

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 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

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

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.




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Cross-references: positive, limit, operator norm, point, proof, theorem

This is version 9 of Fréchet derivative is unique, born on 2006-08-04, modified 2006-10-10.
Object id is 8221, canonical name is FrechetDerivativeIsUnique.
Accessed 1489 times total.

Classification:
AMS MSC46G05 (Functional analysis :: Measures, integration, derivative, holomorphy :: Derivatives)

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