For all $x\not =0 \in\mathbb{R}^m$ and $h\in\mathbb{R}^m$, who is $\xi'(x)\cdot h$?

Question

If $f,g:\mathbb{R}^m\rightarrow\mathbb{R}^n$ are differentiable with $f\in C^1$ and $\xi:\mathbb{R}^m\rightarrow \mathbb{R}$ where $$\xi(x)=\langle g(x),\int_0^{|x|}f(tx)dt\rangle,$$ with $|x|=\sqrt{\langle x,x\rangle}$, then for all $x\not =0 \in\mathbb{R}^m$ and $h\in\mathbb{R}^m$, who is $\xi'(x)\cdot h$?

I don't know how to proceed with the second term. What I've done is:

$\forall x\not =0 \in\mathbb{R}^m$ and $h\in\mathbb{R}^m$,

$$\xi'(x)\cdot h=\lim_{w\to 0}{\frac{\xi(x+wh)-\xi(x)}{w}}=$$ $$=\lim_{w\to 0}\left\langle\frac{g(x+wh)-g(x)}{w},\frac{\int_{0}^{|x+wh|}f(tx+twh)dt-\int_{0}^{|x|}f(tx)dt}{w}\right\rangle=$$ $$=\left\langle g'(x)\cdot h,\lim_{w\to0}\left(\frac{\int_{0}^{|x+wh|}f(tx+twh)dt-\int_{0}^{|x|}f(tx)dt}{w}\right)\right\rangle$$

And after that, how could I establish that $$\lim_{h\to 0}\frac{r_\xi(h)}{|h|}=\lim_{h\to 0}\frac{\xi(x+h)-\xi(x)-\xi'(x)\cdot h}{|h|}=0?$$

Answer 1

Seems something is wrong.

Lets try to calculate the derivative of $\xi$ use inner product linearity.

$$\xi(x+wh)-\xi(x) = \langle g(t+wh), \int_{0}^{|x+wh|}f(t(x+wh))dt \rangle -\langle g(x), \int_{0}^{|x|}f(tx)dt \rangle$$.

Lets add and subtract $\langle g(x), \int_{0}^{|x+wh|}f(t(x+wh))dt\rangle$.

We will get

$$\xi(x+wh)-\xi(x) = \langle g(t+wh), \int_{0}^{|x+wh|}f(t(x+wh))dt \rangle - \langle g(x), \int_{0}^{|x+wh|}f(t(x+wh))dt\rangle+ \langle g(x), \int_{0}^{|x+wh|}f(t(x+wh))dt\rangle-\langle g(x), \int_{0}^{|x|}f(tx)dt \rangle$$

Combining the first two and the last twoyou will have

$$\xi(x+wh)-\xi(x) = \langle g(t+wh)-g(x), \int_{0}^{|x+wh|}f(t(x+wh))dt \rangle + \langle g(x), \int_{0}^{|x+wh|}f(t(x+wh))dt-\int_{0}^{|x|}f(tx)dt \rangle$$.

Dividing by $w$ and taking limit you'll get

$$\xi(x) \cdot h = \lim_{w \to 0} \left(\langle \frac{g(t+wh)-g(x)}{w}, \int_{0}^{|x+wh|}f(t(x+wh))dt \rangle + \langle g(x), \frac{\int_{0}^{|x+wh|}f(t(x+wh))dt-\int_{0}^{|x|}f(tx)dt}{w} \rangle \right)$$.

So we get

$$ \xi'(x) \cdot h = \langle g'(x)\cdot h, \int_{0}^{|x|}f(tx)dt\rangle + \langle g(x), \int_{0}^{|x|}f'(tx)\cdot th \ dt \rangle$$.

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The derivation of the integral part We consider this

$$ \lim_{w \to 0} \left(\langle g(x), \frac{\int_{0}^{|x+wh|}f(t(x+wh))dt-\int_{0}^{|x|}f(tx)dt}{w} \rangle \right)$$

As the first component has no $w$, we let's just look at

$$\lim_{w \to 0} \frac{\int_{0}^{|x+wh|}f(t(x+wh))dt-\int_{0}^{|x|}f(tx)dt}{w} =\lim_{w \to 0} \frac{\int_{0}^{|x+wh|}f(t(x+wh))dt-\int_{0}^{|x|}f(t(x+wh))dt + \int_{0}^{|x|}f(t(x+wh))dt -\int_{0}^{|x|}f(tx)dt}{w} $$

Let's combine first two terms and last two terms.

$$\lim_{w \to 0} \frac{\int_{|x|}^{|x+wh|}f(t(x+wh))dt + \int_{0}^{|x|}(f(t(x+wh))-f(tx))dt}{w} $$

In the second term interval of integration does not depend on $w$, so we can just differentiate under the sign of integration.

The first integral will surely converge to 0 as $w \to 0$, as the region of integration will become a point. But how fast?

As the integral of upper bound We can rewrite

$$\lim_{w \to 0} \frac{\int_{|x|}^{|x+wh|}f(t(x+wh))dt }{w}= f(|x|(x))\cdot h$$.

(So sorry I was wrong in my calculations).

So the correct answer should be

$$ \xi'(x) \cdot h = \langle g'(x)\cdot h, \int_{0}^{|x|}f(tx)dt\rangle + \left\langle g(x), \ f(|x|(x))\cdot h + \int_{0}^{|x|}f'(tx)\cdot th \ dt \right\rangle$$.