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Let $E$,$F$ be Banach spaces,$f\in C^1(E,F)$ is positively homogeneous of degree 1(e.g. $f(tx)=tf(x)$ for $t>0$ and $x\in E\backslash\{0\}$),then $f\in\mathcal{L}(E,F)$.

From \begin{equation*} \underset{t\to1}{\lim}||\frac{f(tx)-f(x)-\partial f(x)(tx-x)}{(t-1)x}||=0 \end{equation*} I obtained $\partial f(x)(x)=f(x)$.But from this,I don't know how to prove $f$ is a bounded linear map.

Note: To say $f$ is differentiable at $x$, if there exists $A\in\mathcal{L}(E,F)$ such that \begin{equation*} \underset{y\to x}{\lim}\frac{f(y)-f(x) - A(y-x)}{||y-x||}=0 \end{equation*}

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  • $\begingroup$ There are several notions of differentiability, Frechet, Gatauex, etc. Which one are you referring to? $\endgroup$ – uniquesolution Apr 15 at 13:07
  • $\begingroup$ I mean Frechet derivative. $\endgroup$ – Tao X Apr 15 at 13:13
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Observe simply that by definition of the Frechet derivative, $\partial f(x)$ is a bounded linear operator from $E$ to $F$, so if you proved that $f(x)=\partial f(x)$, you are almost done. You just need to convince yourself that there is one linear bounded operator that works for all $x$.

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