0
$\begingroup$

Let $F :\mathbb{R}^2 \to \mathbb{R} $ be defined as $F(\vec x)$ =$\Vert \vec x \Vert^2$. Use the definiton of the derivative $\lim_{\vec h \to 0} \frac{||F(\vec x+ \vec h) - F(\vec x)-DF(\vec x)\vec h||}{||\vec h||} = 0$ to show that $F$ is differentiable at every $\vec x \in \mathbb{R}^n, $ and $DF(\vec x)$ = $[2x_1 \quad 2x_2 \quad ... \quad 2x_n ]$ (or $2 \vec x^T$, a row vector)

So I know that $||\vec x^2||$ can be re-written at $|x|^2$ which is the same as $(x)(x)$. Plugging in the value of $\vec x$ into the equation I get $\lim_{\vec h \to 0} \frac{||F(\vec ||x||^2+ \vec h) - F(\vec ||x||^2)-DF(\vec ||x||^2)\vec h||}{||\vec h||} = 0$ which simplifies to $\lim_{\vec h \to 0} \frac{||F(\vec h) -DF((2) (\vec||x||))\vec h||}{||\vec h||} = 0$, Which I assume simplifies to $\lim_{\vec h \to 0} \frac{||DF((2)\vec||x||}{||\vec h||} = 0$ ?

Not sure if I am correct and I am having trouble proving that $DF(\vec x)$ is a row vector.

Any help is appreciated!

1 Answers 1

0

Let $L(x) = 2x^T$ (which we must prove that it equals the differential of $f$). Then,

$$|f(x+h)-f(x) - L(x)\cdot h| = |\|x+h\|^2 - \|x\|^2 - 2x \cdot h| = |\|x\|^2 +\|h\|^2 + 2 x\cdot h - 2 x\cdot h| = \|h\|^2$$

Edit:

Hence:

$$\lim_{h\to 0} \frac{|f(x+h)-f(x) - L(x)\cdot h| }{\|h\|} = \lim_{h\to 0} \|h\| = 0$$

By definition, this means that $f$ is differentiable at $x$ and that $Df(x) = L(x)$. In other words:

$$Df(x) = 2 x^T = 2 [x_1 \cdots x_n]$$

  • 0
    Is that the proof for the row vector? What about showing that $F$ is differentiable at all $\vec x$ ? – 2017-01-26
  • 0
    @quwerty well, $x$ was arbitrary, so this does show that it is differentiable at all $x$. – 2017-01-26
  • 0
    So if I understand what you've done, you have shown that the absolute value of the derivative is the norm of $h^2$? And thus it is differentiable? I don't see how it can be a row vector though. – 2017-01-26
  • 0
    @quwerty I hope it's clear now. – 2017-01-26
  • 0
    Ahh yes it makes perfect sense now. Thank you :) – 2017-01-26