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Let $E$ and $F$ be Banach spaces and let $X \subset E$ be open. Call maps $\phi,\psi:X\rightarrow F\;$ tangent at the point $p \in X$ if $$ \lim_{x \to p} = \frac{|\phi(x)-\psi(x)|_F}{|x-p|_E}=0 $$ For a given map $f:X\rightarrow F$, I am trying to show that there is at most one bounded linear map $L:E\rightarrow F$ that is tangent to the map $g_L(x) = f(p) + L(x - p)$. Demonstrating this claim will show that the derivative is unique. So, suppose there exists two such maps $L_1$ and $L_2$. Applying the triangle inequality and properties of the operator norm of bounded linear functions, $$ 0 = \lim_{x \to p}\frac{1}{|x - p|}\left( |f(x) - f(p) - L_1(x - p)| - |f(x) - f(p) - L_2(x - p)| \right) $$ $$ \leq \lim_{x \to p}\frac{1}{|x - p|}\left( |f(x) - f(p)| + |L_1(x-p)| - |f(x) - f(p)| - |L_2(x-p)| \right) $$

$$ = \lim_{x \to p}\frac{1}{|x - p|} \left(|L_1(x-p)| - |L_2(x-p)|\right) $$ $$ \leq \lim_{x \to p}\frac{1}{|x - p|} \left(|L_1|\cdot |x-p| - |L_2|\cdot |x-p|\right) $$ $$ = |L_1| - |L_2| $$ So, $$ 0 \leq |L_1| - |L_2| \implies |L_1| \geq |L_2|. $$ By a symmetrical argument $|L_2| \geq |L_1|$ and therefore, $|L_1| = |L_2|$.

So, my question is, am I on the right track? I believe my proof correctly shows that $|L_1| = |L_2|$, however, this does not imply that $L_1 = L_2$. Can this approach be salvaged?

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I don't know if you are in the right track, but I think there is a problem, since in the best case you will manage to show something only about $p$. –  Davide Giraudo Jan 25 '12 at 22:38

2 Answers 2

up vote 2 down vote accepted

The first inequality is not justified, because $-|f(x)-f(p)-L_2(x-p)|\geq -|f(x)-f(p)|-|L_2(x-p)|$ rather than the other way around. Also, there is no justification for the limits in the second and third lines to exist. (In particular, note that $\lim\limits_{x\to p} \frac{|L_1(x-p)|}{|x-p|}$ usually does not exist for a linear map $L_1$.) And if the limit in the third line does exist, the subsequent inequality isn't justified, because $-|L_2(x-p)|\geq -|L_2||x-p|$ rather than the other way around.

Rather, the following approach could be used. For each $x\neq p$, the triangle inequality implies that $$\begin{align*} \frac{|(L_1-L_2)(x-p)|}{|x-p|}&=\frac{|L_1(x-p)-L_2(x-p)|}{|x-p|}\\ &\leq \frac{|f(x)-f(p)-L_1(x-p)|}{|x-p|}+\frac{|f(x)-f(p)-L_2(x-p)|}{|x-p|}. \end{align*}$$ Since the hypothesis is that each of the fractions on the right-hand side goes to $0$ as $x\to p$, it follows that for all $\varepsilon>0$, there exists $\delta>0$ such that $|(L_1-L_2)(x-p)|\leq \varepsilon |x-p|$ whenever $0<|x-p|<\delta$. Therefore $|L_1-L_2|=\sup\limits_{0<|y|<\delta}\frac{|(L_1-L_2)(y)|}{|y|}\leq \varepsilon$. Since $\varepsilon>0$ was arbitrary, this shows that $L_1=L_2$.

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In fact, we can copy the proof of the uniqueness of the differential of a function. Fix $h\neq 0$. Then we should have $$\lim_{t\to 0^+}\frac{|f(p)+L(th+p)-L_j(th+p)|}{t|h|}=0\quad \forall j\in\{1,2\},$$ so $\lim_{t\to 0^+}\frac{L_1(th+p)-L_2(th+p)}{t|h|}=0$ and so $$\lim_{t\to 0^+}\frac{(L_1-L_2)h+(L_1-L_2)\left(\frac pt\right)}{|h|}=0,$$ hence denoting $L'=L_1-L_2$: $\lim_{t\to 0^+}\frac 1t|L'(p)|=|L'(h)|$ for all $h$, hence $L'(p)=0$ and $L'(h)=0$. Since $h$ was arbitrary, $L_1=L_2$.

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