Vrftsjtryk

Let $E$ be a normed space and $Omega subset E$ be an open convex subset. Let $ain Omega$ and $f:Omegalongrightarrow mathbb{R}$, we say $f$ is differentiable in the direction $v$ at $a$ if the following limit exists and finite
$$ f'(a,v):=lim_{tlongrightarrow 0} frac{f(a+tv) - f(a)}{t}.$$
My question is, which of the following should be the correct definition of Gateaux differentiability at $a$?

1. 
   $f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear.


2. 
   $f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear and continuous.

The question I have in mind is the following

Let $Omegasubset E$ be open, convex and
$f:Omegalongrightarrowmathbb{R}$ be convex. Let $ain Omega$, if
$f'(a,v)$ exists and finite for all $vin E$ then $f$ is Gateaux
differentiable at $a$.

If I use the first definition then it is obvious, but for the second definition then I don't see how it is true. Since we need $f$ to be bounded or continuous (to get $f$ is Lipschitz near $a$) in order to show $vlongmapsto f'(a,v)$ is continuous.

Proof for $vlongmapsto f'(a,v)$ is linear given that $f$ is convex and $f'(a,v)$ exists for all $v$.

First of all given this hypothesis I can show that
$$ f'_+(a,v):=lim_{tlongrightarrow 0^+} frac{f(a+tv) - f(a)}{t}.$$
exists and finite everywhere. Indeed, the function $varphi(s) = f(a+sv)$ for $sin (-varepsilon,varepsilon)$ is convex from a subset of $mathbb{R}longrightarrow mathbb{R}$, hence it is continuous and locally Lipschitz. By the convexity $slongmapstofrac{f(a+sv) - f(a)}{s}$ is decreasing as $slongrightarrow 0^+$. Together with being bounded from
$$frac{f(a+sv) - f(a)}{s} = frac{varphi(s) - varphi(0)}{s} geq -frac{Cs}{s} = -C.$$
we conclude that $f'_+(a,v)$ exists and finite for all $v$. It is clear that
$$f'_+(a,lambda v) = lambda f'_+(a,v).$$

Let $p(v) = f'_+(a,v)$, it is easy to see that $p:Elongrightarrow mathbb{R}$ is sublinear since for $u,vin E$, we have
begin{align*}
p(u+v) &= lim_{tlongrightarrow 0^+} frac{f(a+t(u+v))-f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)+f(a+2tv) -f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)-f(a)}{2t} + lim_{tlongrightarrow 0^+} frac{f(a+2tv)-f(a)}{2t} = p(u) + p(v).
end{align*}
and $0leq p(0) leq p(-v)+p(v)$, thus $-p(-v)leq p(v)$ for $vin E$.
Now let us define
$$V = {vin E: -p(-v) = p(v)} = {vin E: f'(a,v);text{exists and finite}}.$$
It is easy to see that $V$ is a linear subspace of $E$. For $u,vin V$ we have
begin{equation*}
-p(-u)-p(-v)leq -p(-u-v) leq p(u+v)leq p(u+v) = -p(-u)+p(-v).
end{equation*}
Therefore $u+vin V$. For $lambda in mathbb{R}$ and $vin V$, if $lambda >0$ then clearly $lambda vin V$, while if $lambda < 0$ then $-lambda v > 0$, thus
$$ p(-lambda v) = -lambda p(v) qquad Longrightarrow qquad -p(-lambda v) = p(lambda v)$$
Hence $lambda v in V$. From that definition it is obvious that $p|_V$ is linear.

If $f'(a,v)$ exists for all $vin E$ then $V = E$, hence $vlongmapsto f'(a,v) = p(v)$ is linear on $E$.

Can we show $vlongmapsto f'(a,v)$ is indeed continuous?

The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.