yesterday I took a test and I can't answer this question. “Let $E,F$ be two normative spaces and f be a function on E to F with Gateaux differential and all limits $$\lim_{t \to 0}\dfrac{f(x+tv)-f(x)}{t}$$ It exists uniformly for all $v \in S_E(0,1)$ then f is Frechet diff. “
My teacher told me that if you understand compactness in $S_E(0,1)$ you can solve that, but I don't know what it means. Thanks for any help
I don't think compactness is needed here. In fact, the result should hold for arbitrary normed spaces (whether they have finite or infinite dimension). The uniform assumption means that there is a bounded linear map $A_x$ such that for all $\varepsilon>0$, there exists $\delta>0$ such that $$\sup_{v\in S(0;1)}\Big|\frac{f(x+tv)-f(x)}{t}-A_xv\Big|_F<\varepsilon$$ whenever $0<|t|<\delta$.
Define $r(x;h)=f(x+h)-f(x)+A_xh$. Let $v_h=\frac{1}{|h|_E}h$. Then, if $|h|_E<\delta$ we have that \begin{align} \frac{|r(x;h)|_F}{|h|_E}&=\frac{|f(x+h)-f(x)-A_xh|_F}{|h|_E}\\ &=\Big|\frac{f(x+|h|v_h)-f(x)}{|h|_E}-A_xv_h\Big|_F<\varepsilon \end{align} This means that $f$ is (Frechet) differentiable at $x$ and that $f'(x)=A_x$.