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Let $E, F$ normed spaces and $f:A\subseteq E\to F$ with $A$ open set, suppose that $f$ is differentiable at $a\in A$ and that $f$ is locally Lipschitz of constant $k>0$ in $a$. Show that $||Df(a)||\leq k$.

Note: $f$ is differentiable in $a$ iff exist $Df(a)\in \mathcal{L}(E, F)$ so that $$\lim_{h\to 0}\frac{||f(a+h)-f(a)-Df(a)h||_F}{||h||_E}=0$$

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If $f$ is locally Lipschitz of constant $k > 0$ about the point $a \in A$ then we have a neighborhood $U \subset A$ for which $$\|f(x) - f(y)\|_F \le k \|x-y\|_E.$$

Notice also that $$\frac{|\|f(a+h) - f(a)\|_F - \|Df(a)h\|_F|}{\|h\|_E} \le \frac{\|f(a+h) - f(a) - Df(a)h\|_F}{\|h\|_E}$$ for all $h$. By the definition of differentiability you have provided, this means that as $h \to 0$ we have $$\frac{\|Df(a)h\|_F}{\|h\|_E} \to \frac{\|f(a+h) - f(a)\|_F}{\|h\|_E} \le k.$$

The inequality is from the Lipschitz property, since for small enough $h$ we know that $a+h$ is in $U$. This gives $$\left\| Df(a) \frac{h}{\|h\|} \right\| < k$$ for all $h$ sufficiently small. However, since $\frac{h}{\|h\|}$ is a unit vector for any $h$, this implies that the operator norm of $Df(a)$ is less than $k$.

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