Given a function $F$, how to evaluate $\max_{t\in (0,1)}\left\vert \frac{\partial^2 F}{\partial x_i \partial x_j}(ty)\right\vert$? As an "extra problem" for my calculus 2 test class we got the following. Since it is an extra problem, I expected it to be more difficult than the other exercises, but actually but not that much. This is the problem:
Let $F:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}$ such that
$$F:(x_1, x_2, x_3)\mapsto -\frac{1}{|x|^2} +\frac{1}{|x|}\in\mathbb{R}.$$
For a fixed $y\in\mathbb{R}^3\setminus\{0\}$, find
$$\max_{t\in (0,1)}\left\vert \frac{\partial^2 F}{\partial x_i \partial x_j}(ty)\right\vert.$$
To start I computed $\nabla F(x)$ which is
$$\nabla F(x) = \left(\frac{2}{|x|^4}-\frac{1}{|x|^3}\right) x,$$
where $x$ denotes the vector $x=(x_1, x_2, x_3)$.
I went ahead by computing the second partial derivatives. I had
$$\frac{\partial^2 F}{\partial x_i^2} = \frac{2}{|x|^4}-\frac{1}{|x|^3} +\frac{3x_i^2}{|x|^5}-\frac{8x_i^2}{|x|^6}, \quad i\in\{1, 2, 3\}$$
and
$$\frac{\partial^2 F}{\partial x_i\partial x_j} = \frac{3x_i x_j}{|x|^5}-\frac{8x_i x_j}{|x|^6}, \quad i\in\{1, 2, 3\}, i\neq j.$$
I got stuck at this point since I have no idea how to establish the maximum. I’ve been thinking about it unsuccessfully for a couple of days, but I am really interested in the solution.
Could someone please help in finding that maximum? Any hint will be very helpful.
Thank you in advance.
 A: Consider the family of functions
$$G_{ij}(t;y):=\frac{\partial^2 F}{\partial x_i\partial x_j}(ty)$$
Write
$$F(x)=f(|x|) \\ f(s)=-\frac{1}{s^2}+\frac{1}{s}$$
Using the chain rule,
$$\frac{\partial F}{\partial x_i}(x)=f'(|x|)~\frac{x_i}{|x|}$$
And the product rule, $$ \frac{\partial^2 F}{\partial x_i\partial x_j}(x)=f''(|x|)\frac{x_ix_j}{|x|^2}+f'(|x|)\frac{\frac{\partial x_i}{\partial x_j}|x|-\frac{x_j}{|x|}x_i}{|x|^2} \\ =f''(|x|)\frac{x_ix_j}{|x|^2}-f'(|x|)\frac{x_ix_j}{|x|^3}+\delta_{ij}f'(|x|)\frac{1}{|x|}$$
Now, compute
$$f'(s)=\frac{2}{s^3}-\frac{1}{s^2} \\ f''(s)=\frac{-6}{s^4}+\frac{2}{s^3}$$
Hence
$$\frac{\partial ^2 F}{\partial x_i\partial x_j}(x)=\frac{x_ix_j}{|x|^2}\left(f''(|x|)-\frac{f'(|x|)}{x}+\delta_{ij}f'(|x|)\frac{1}{|x|}\right) \\ 
=\frac{x_ix_j}{|x|^2}\left(\frac{-6}{|x|^4}+\frac{2}{|x|^3}-\frac{2}{|x|^4}+\frac{1}{|x|^3}+\delta_{ij}\left(\frac{2}{|x|^4}-\frac{1}{|x|^3}\right)\right) \\ =\frac{x_ix_j}{|x|^5}\left(3-\frac{8}{|x|}+\delta_{ij}\left(\frac{2}{|x|}-1\right)\right) \\ 
=\frac{x_ix_j}{|x|^5}\left((3-\delta_{ij})+\frac{2}{|x|}\left(\delta_{ij}-4\right)\right)$$
Therefore
$$G_{ij}(t;y)=\frac{ty_ity_j}{|ty|^5}\left((3-\delta_{ij})+\frac{2}{|ty|}\left(\delta_{ij}-4\right)\right) \\ =|t|^{-3}\frac{y_iy_j}{|y|}\left((3-\delta_{ij})+t^{-1}\frac{2}{|y|}\left(\delta_{ij}-4\right)\right)$$
So for $i\neq j$ and $t\in(0,1)$,
$$|G_{ij}(t;y)|=b~t^{-3}|3-8at^{-1}|$$
Where $a=|y_iy_j/|y|^5|$ and $a=1/|y|$. This clearly is unbounded.
On the other hand for $i=j$,
$$|G_{ii}(t;y)|=bt^{-3}|2-6at^{-1}|$$
Same thing.
