1

Assume: $f:R^d \to R^m$ is a function, so $c>0$ and $\alpha\gt1$ exists with $||f(x)||\le c||x||^\alpha$ for all x in the neighbourhood of $0\in R^d$.

Show that $f$ is differentiable at $0$ and Calculate the Derivation

I made this: With the Formula $$lim_{x\to x_0} \frac{f(x)-f(x_0)-A(x-x_0)}{||x-x_0||} = 0$$ comes $$\lim_{x\to 0} \frac{f(x)-f(0)-A(x)}{||x||} =0$$ and with Definition $$lim_{x\to0}\frac{c||x||^\alpha - A(x)}{||x||}=0$$ so $$\lim_{x\to0}\frac{A(x)}{||x||}=\frac{c||x||^\alpha}{||x||}=c||x||^{\alpha-1}$$

but Derivation of $c||x||^\alpha$ is $\alpha c||x||^{\alpha-^1}$ isn't it?

It seems i made a mistake. Can someone revise it please

Thanks

suugii19
  • 69

1 Answers1

0

We have $\|f(0)\| \le c \|0\|^{\alpha} = 0$, so $f(0) = 0$. Now, for $h$ in a neighbourhood of $0$:

$$\frac{\|f(0+h) - f(0) - 0\|}{\|h\|} =\frac{ \|f(h)\|}{\|h\|} \le \frac{c\|h\|^{\alpha}}{\|h\|} = c \|h\|^{\alpha-1}$$

which goes to $0$ as $h \to 0$. Hence, $f$ is Fréchet-differentiable at $0$ and $Df(0) \equiv 0$.