Reference:http://en.m.wikipedia.org/wiki/Gateaux_derivative ('Linearity and continuity' section)
Let $V,W$ be Banach spaces over $\mathbb{K}$ and $E$ be open in $V$.
Let $f:E\rightarrow W$ be a function which is Gateaux-differentiable at $p$. That is, $\forall h\in V, \lim_{t\to 0}\frac{f(p+th)-f(p)}{t}$ exists.
Let $df(p;h)$ denote the Gateaux-differential of $f$ at $p$ in the direction $h$.
Define $F(h)=df(p;h)$ for all $h\in V$.
How do I prove that $F$ is linear?
Here's how I tried:
Fix $h,k\in V$ and define $g(t,s)=f(p+th+sk)$ where $s,t$ are (real)complex numbers. Then, $g_t(0,0)=df(p;h)$ and $g_s(0,0)=df(p;k)$ and $\lim_{t\to 0} \frac{g(t,t)-g(0,0)}{t}=df(p;h+k)$. In general, $g$ is differentiable along any straight line passing through the origin.
Since I don't know the continuity of $g$, I cannot find a relation among the above equalities..
Please help.. Thank you in advance
DIFFERENTIATING AT THE ORIGIN.
Take the vector space to be the ordinary $(x,y)$ plane. In polar coordinates, restricting to $r \geq 0,$ let $$ f(r, \theta) = r \sin 3 \theta . $$ A line through the origin is given by fixing a value of $\theta;$ doing so, the Gateaux ratio is just the constant $\sin 3 \theta ,$ so that is also the limit.
The Gateaux ratio is $0$ in linearly independent directions, angles $0,\pi/3. $ If it were linear, we would always get the Gateaux ratio being $0.$ But that is not the case. So, not linear.
Oh, we need the $3$ to be odd to get correct comparison between $\theta$ and $\theta + \pi.$ If we replaced the $3$ by $1,$ we would have $f = y,$ differentiable in every sense.