0
$\begingroup$

Let $U\subset\mathbb R^m$ be an open subset.

I would like to prove the following equivalence:

$f:U\to \mathbb R^n$ is differentiable at the point $a\in U$

$\Leftrightarrow$

There is for each $h\in \mathbb R^m$ with $a+h\in U$, a linear transformation $A(h):\mathbb R^m\to \mathbb R^n$ such that $f(a+h)-f(a)=A(h)\cdot h$ and $h\mapsto A(h)$ is continuous at $h=0$.

I've already tried to prove both sides of this equivalence without success. I need a hint to tackle this question.

1 Answers 1

1

Here is a sketch:

In the forward direction, you know there is some $A$ such that $f(x+h)-f(x) -Gh = r(h)$, where $r$ is $o(h)$, so you need only define a suitable $A(h)$. Let $\Delta(0) = 0$ and $\Delta(h) = r(h) {h^T \over \|h\|^2} $. Let $A(h) = G + \Delta(h)$ to finish, check that $A(\cdot)$ is continuous at $h=0$. (Note that you can define $A(\cdot)$ in other ways, this is a straightforward way.)

In the reverse direction, write $f(x+h)-f(x) = A(0)h + (A(h)-A(0)) h$ and show that $r(h) = (A(h)-A(0)) h$ is $o(h)$.

  • 0
    In the forth line do you mean $A(h)=A(h)+\Delta(h)$? thank you for your answer – 2017-01-06
  • 0
    No, that would mean that $\Delta(h) = 0$. $A$ is the derivative of $f$, and I am defining $A(h)$. Let me change the symbols to avoid confusion. – 2017-01-06
  • 0
    Thank you again. What's this $T$? – 2017-01-06
  • 0
    Do you mean the transpose on $h^T$? – 2017-01-06
  • 0
    Of course! my mistake – 2017-01-06
  • 0
    Can I replace $h^T$ with the inner product? – 2017-01-06
  • 0
    If your inner product is $\langle a,b \rangle = a^Tb$ then yes... – 2017-01-06