If the Gateaux derivative is not linear does this mean the Frechet derivative doesn't exist? It seems to be true as if the frechet exists then it is the same as the gateaux which would then be nonlinear (contradiction) but this seems useful yet I can't really find it anywhere and the lecturer hasn't mentioned it. Thanks.
 A: I don't when if you are working in Euclidean spaces or normed spaces, but
anyway, Frechet differentiability at a point $x_{0}$ for a function
$f:E\rightarrow Y$, where $E\subseteq X$, means
$$
\lim_{x\rightarrow x_{0}}\frac{\Vert f(x)-f(x_{0})-L(x-x_{0})||_{Y}}{\Vert
x-x_{0}||_{X}}=0
$$
where $L:X\rightarrow Y$ is linear and continuous. Now if $x_{0}$ is an
interior point of $E$, then taking a direction $v\in X$ with norm $\Vert
v||_{X}>0$ and $x=x_{0}+tv$, you get that
\begin{align*}
0  & =\lim_{x\rightarrow x_{0}}\frac{\Vert f(x)-f(x_{0})-L(x-x_{0})||_{Y}%
}{\Vert x-x_{0}||_{X}}=\lim_{t\rightarrow0}\frac{\Vert f(x_{0}+tv)-f(x_{0}%
)-L(x_{0}+tv-x_{0})||_{Y}}{\Vert x_{0}+tv-x_{0}||_{X}}\\
& =\lim_{t\rightarrow0}\frac{\Vert f(x_{0}+tv)-f(x_{0})-tL(v)||_{Y}}{|t|\Vert
v||_{X}}=\frac{1}{\Vert v||_{X}}\lim_{t\rightarrow0}\left\Vert \frac
{f(x_{0}+tv)-f(x_{0})}{t}-L(v)\right\Vert ,
\end{align*}
which implies that there exists the Gateaux derivative in the direction $v$
with$$
\frac{\partial f}{\partial v}(x_{0})=L(v).
$$
Since $L$ is linear, it follows that $v\mapsto\frac{\partial f}{\partial
v}(x_{0})$ has to be linear.
