Geometric connection between Dini derivative and scalar products I am reading through a paper (Sec. 4.5) that uses a geometric construction not quite clear to me.
Namely, let $B$ be a closed ball in $\mathbb R^n$ centered at $x$.
Let $s \in B$.
Next, let $w$ be an arbitrary vector.
Then, for a sufficiently small $\varepsilon > 0$, there is $z \in \partial B$ such that
$$ z - (s + \varepsilon w) $$
is parallel to $x-s$ and pointing in the same direction.
That is,
$$
\Biggl\langle z - (s + \varepsilon w), \frac{x-s}{\| x - s \|} \Biggr\rangle = \| z - (s + \varepsilon w) \|
$$
While I understand that such a $z$ can be found, I fail to see why
$$
\Biggl\langle z - s, \frac{x-s}{\| x - s \|} \Biggr\rangle = \mathcal O(\varepsilon^2)
$$
The paper then claims that
$$
- \varepsilon \Biggl\langle w, \frac{x-s}{\| x - s \|} \Biggr\rangle = \| z - (s + \varepsilon w) \| + \mathcal O(\varepsilon^2)
$$
which clearly uses the above statement.
I can visualize how to find $z$ so that $z - (s + \varepsilon w)$ and $x-s$ are parallel.
But $\varepsilon$ is small so that the direction of the vector $s$ is barely changed.
Then, it seems $z-s$ is almost parallel to $x-s$.
Why should their scalar product be small?

This would be correct if $s$ were itself in $\partial B$. The paper doesn’t say it is, but perhaps there’s a reason to assume this? Apparently $s$ is a point where $V$ is minimal in $B$ (section $4.2$, item $2$). Perhaps it can be shown that the minimum must be attained on the boundary? That’s the only thing I can think of that would make sense of this.
