Fréchet derivative is unique

TheoremThe 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.