Strictly differentiable functions 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 $U$.
 A: ($\Rightarrow$): 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.
