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Let $E, F$ be normed vector spaces over $\mathbb{R}$. Let $U$ be an open subset of $E$. Let $a \in U$. Let $f\colon U \rightarrow F$ be a map. Suppose there exists a continuous linear map $L\colon E \rightarrow F$ such that $\frac {|f(x) - f(y) - L(x - y)|}{|x - y|} \rightarrow 0$ when $(x, y) \rightarrow (a, a)$. Then $f$ is called strictly differentiable at $a$.

Is the following proposition true? If yes, how do we prove it?

Proposition Let $E, F, U, f$ be as above. Then $f$ is strictly differentiable at every point of $U$ if and only if $f$ is of class $C^1$ on $U$.

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    @copper.hat That's$a$part of the question, not the assumption.2012-10-22

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($\Rightarrow$): Suppose for all $a \in U$, $ \epsilon >0$, there exists $L_a$, $\delta>0$ such that whenever $x,y \in B(a,\delta)$, $\|f(x)-f(y)-L_a(x-y)\| \leq \|x-y\|$. Then by choosing $y=a$, it is clear that $f$ is Fréchet differentiable at $a$, and $Df(a) = L_a$. It remains to be shown that the derivative is continuous.

Choose $y \in B(a,\frac{\delta}{3})$, and note that $B(y,\frac{\delta}{2}) \subset B(a,\delta)$. By hypotheses, $\exists \delta' >0$ (with $\delta'\leq \frac{\delta}{2}$, without loss of generality) such that if $x \in B(y,\delta')$ then $\|f(x)-f(y)-L_y(x-y)\| \leq \epsilon \|x-y\|$. Letting $h=x-y$ we have $\|L_y h-L_a h\| = \|f(y+h)-f(y)-L_a h-[f(y+h)-f(y)-L_y h]\| \leq 2 \epsilon \|h\|$. Since this holds for all $\|h\|< \delta'$, we have $\|L_y-L_a\| \leq 2 \epsilon$, from which it follows that $a \mapsto L_a = Df(a)$ is continuous, and hence $f \in C^1(U)$.

($\Leftarrow$): Suppose $f \in C^1(U)$. We need an estimate, which is similar the mean value theorem. Choose $a \in U$ and $\epsilon>0$. Choose $\delta>0$ such that $B(a,\delta)\subset U$ and $\sup_{x \in B(a, \delta)} \|Df(x)-Df(a)\| < \epsilon$. Now let $\lambda \in F^*$, then using the ordinary mean value theorem, we have $\lambda(f(x))-\lambda(f(y)) = \int_0^1 \lambda(Df(y+t(x-y))(x-y)) dt$. Subtracting $\lambda(Df(a)(x-y))$ from both sides gives $\lambda(f(x)-f(y)-Df(a)(x-y)) = \int_0^1 \lambda([Df(y+t(x-y))-Df(a)](x-y)) dt$ Thus $|\lambda(f(x)-f(y)-Df(a)(x-y))| \leq \epsilon \|\lambda\| \|x-y\|$ Since this is true for all $\lambda \in F^*$, it follows (using the Hahn Banach theorem) that $|f(x)-f(y)-Df(a)(x-y)| \leq \epsilon \|x-y\|$. Letting $L_a = Df(a)$ we see that $f$ is strictly differentiable.

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    Thanks. For the readers who don't have the Rudin's book. Let $E$ be$a$normed vector space over $\mathbb{R}$ or $\mathbb{C}$, $a \in E, a \neq 0$. Then there exists $f \in E^*$ such that $||f|| = 1$ and $f(a) = ||a||$. The proof is as follows. Let $M = \{\lambda a\}$ be the subspace of $E$ spanned by $a$. Define $g(\lambda a) = \lambda ||a||$. Then $g \in M^*$ and $||g|| = 1$. Then apply the Hahn-Banach to $g$.2012-10-23