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.
If the Gateaux derivative is not linear does this mean the Frechet derivative doesn't exist?
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real-analysis
derivatives
1 Answers
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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.