Show that if Frechet (d) derivative is exist this implies that Gateaux derivative is exist.

Question

Show that if Frechet (d) derivative is exist this implies that Gateaux (D) derivative is exist , and D f(x)= d f(x), and show that the opposite is not true always and why?

my solution

suppose that the f-derivative exist

$\lim_{\| k \| \to 0} \frac{\| T(x+k)-T(x)-dT(x)k)\| }{\| k\|}=0 \quad Put \quad k=\alpha h \rightarrow $ $\lim_{\| \alpha \| \to 0} \frac{ \| T(x+\alpha h-T(x)-dT(x)\alpha h\|}{\|\alpha h \|}=0 $

By tailor series:

$ T(x+\alpha h)=T(x)+\alpha dT(x)h+h O(\alpha) \leftrightarrow T(x+\alpha h)-T(x)=\alpha dT(x)h+h O(\alpha)$

$\lim_{\alpha \to 0} \frac{T(x+\alpha h)-T(x)-dT(x)\alpha h}{\alpha} = 0$

$\lim_{\alpha \to 0} \frac{T(x+\alpha h)-T(x)}{\alpha} -dT(x)h = 0$

$\lim_{\alpha \to 0} \frac{T(x+\alpha h)-T(x)}{\alpha} =dT(x)h$

and i can't do second term of question "and show that the opposite is not true always and why?" .

Answer 1

Consider the function $f\colon \mathbf R^2 \rightarrow \mathbf R$ defined by $f(0,0)=0,\enspace f(x,y) = \dfrac{xy^3}{x^2+y^6}$. It is Gâteaux-derivable at $(0,0)$, but not Fréchet-derivable. Indeed, if it were Fréchet-derivable, it should be continuous at $(0,0)$. Hint: let $(x,y) \to (0,0)$ along the curve $x=y^3$.