Fréchet derivative is unique

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

$$\parallel f(𝐱+𝐡)-f(𝐱)-A𝐡\parallel \le \frac{\epsilon }{2}\cdot \parallel 𝐡\parallel \text{ and }\parallel f(𝐱+𝐡)-f(𝐱)-B𝐡\parallel \le \frac{\epsilon }{2}\cdot \parallel 𝐡\parallel$$

holds. This gives

$\parallel (A-B)𝐡\parallel$ $=\parallel (f(𝐱+𝐡)-f(𝐱)-A𝐡)-(f(𝐱+𝐡)-f(𝐱)-B𝐡)\parallel$

$\le \parallel f(𝐱+𝐡)-f(𝐱)-A𝐡\parallel +\parallel f(𝐱+𝐡)-f(𝐱)-B𝐡\parallel$

Now we have

$$\delta \cdot \parallel A-B\parallel =\delta \cdot \underset{\parallel 𝐠\parallel \le 1}{sup}\parallel (A-B)𝐠\parallel =\underset{\parallel 𝐠\parallel \le \delta }{sup}\parallel (A-B)𝐠\parallel \le \underset{\parallel 𝐠\parallel \le \delta }{sup}\epsilon \cdot \parallel 𝐠\parallel \le \epsilon \cdot \delta ,$$

thus $\parallel A-B\parallel \le \epsilon$ as we wanted to show.

Title	Fréchet derivative is unique
Canonical name	FrechetDerivativeIsUnique
Date of creation	2013-03-22 16:08:35
Last modified on	2013-03-22 16:08:35
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	12
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 46G05
Related topic	derivative