How are these definitions of Frechet derivatives related to each other? First def. for f exists the fréchet-derivative if there is a continuous operator $T \in L(X,Y)$, such that $\lim_{h \rightarrow 0} \frac{f(x_0 + hv) - f(x_0)}{h} = Tv$ converges uniformly in $v$, vs.
Second def. $f(x_0+h)=f(x_0)+ Th + o(h)$.
How do get the uniform convergence from the second one? I guess it is pretty obvious, but I do not see it.
My doubt is, that this definition might not be true for every normed vector space, since not every norm is equivalent to the sup-norm that refers to uniform convergence. Or does uniform convergence for a abstract normed vector space refer to something different than convergence in the sup-norm?
 A: The "uniform convergence" of the first definition is not uniform convergence on all of $X$. If $f$ is defined only on a proper subset of $X$, it cannot possibly mean that since $f(x_0 + hv)$ isn't defined for all $v$, and even if it is defined on all of $X$, the difference quotient
$$\frac{f(x_0 + hv) - f(x_0)}{h}$$
has no reason whatsoever to be close to $Tv$ when $\lVert v\rVert$ is large, really really large.
What is meant is uniform convergence on bounded subsets of $X$. It is enough to pick one special bounded subset of $X$, the (closed) unit ball $B_X = \{ v\in X : \lVert v\rVert \leqslant 1\}$.
Then the first definition means that the function
$$q(h) := \sup_{v \in B_X} \left\lVert \frac{f(x_0 + hv) - f(x_0)}{h} - Tv\right\rVert\tag{1}$$
has the property that $\lim\limits_{h\downarrow 0} q(h) = 0$.
The second definition means that
$$r(w) := f(x_0+w) - f(x_0) - Aw$$
belongs to $o(w)$, which in turn means
$$\lim_{w\to 0} \frac{r(w)}{\lVert w\rVert} = 0,$$
and that means that for every $\varepsilon > 0$, there is a $\delta > 0$ such that
$$\lVert w\rVert < \delta \Rightarrow \lVert r(w)\rVert \leqslant \varepsilon\cdot \lVert w\rVert,$$
or, with $w_0 = \frac{1}{\lVert w\rVert}\cdot w$ for $w \neq 0$,
$$\sup_{\lVert w_0\rVert = 1,\, h < \delta} \left\lVert \frac{f(x_0 + hw_0) - f(x_0)}{h} - Aw_0\right\rVert \leqslant \varepsilon.\tag{2}$$
Now, if we write $v = t\cdot v_0$ with $\lVert v_0\rVert = 1$ and $t\in [0,1]$ in $(1)$, we see that
$$\begin{align}
\left\lVert \frac{f(x_0 + htv_0)-f(x_0)}{h} - T(tv_0)\right\rVert &= t\cdot \left\lVert \frac{f(x_0+htv_0) - f(x_0)}{th} - Tv_0\right\rVert\\
&\leqslant \left\lVert \frac{f(x_0+htv_0) - f(x_0)}{th} - Tv_0\right\rVert,
\end{align}$$
so we have
$$\sup_{h < \delta} \sup_{\lVert v\rVert \leqslant 1}\left\lVert \frac{f(x_0+hv) - f(x_0)}{h} - Tv\right\rVert = \sup_{h < \delta} \sup_{\lVert v\rVert = 1}\left\lVert \frac{f(x_0+hv) - f(x_0)}{h} - Tv\right\rVert,\tag{3}$$
and $\lim\limits_{h\downarrow 0} q(h) = 0$ is equivalent to the second definition, since with regard to $(3)$, it reduces to $(2)$.
