# $f\in C^1(E,F)$ is positively homogeneous of degree 1,then $f\in\mathcal{L}(E,F)$

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 $$\begin{equation*} \underset{y\to x}{\lim}\frac{f(y)-f(x) - A(y-x)}{||y-x||}=0 \end{equation*}$$

• There are several notions of differentiability, Frechet, Gatauex, etc. Which one are you referring to? – uniquesolution Apr 15 at 13:07
• I mean Frechet derivative. – Tao X Apr 15 at 13:13

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 $$x$$.