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?