Let $f:\mathbb{R}^p \to \mathbb{R}$.If $D_uf$ exists for a unit vector $u \in \mathbb{R}^p$, then show that $$- D_u f = D_{-u} f$$
If $f$ was differentiable, the directional derivative in the direction of $u$, would be $$\nabla f \cdot u,$$ so from that we could have shown the result easily.However, since it is not the case, I directly tried to use the definition of directional derivative, i.e I have assumed that $$lim_{h\to 0} f(c+ hu) - f(c) / h, \quad \forall c\in \mathbb{R}^p,$$ exists, and we can rewrite it as $$lim_{h\to 0^+} \frac{f(c + |h|c) - f(c)}{|h|} = lim_{h\to 0^-} \frac{-f(c - |h|u) + f(c)}{|h|}. \quad (1)$$ Then define t = |h|, hence $$lim_{t\to 0^+} \frac{f(c+ tu) - f(c)}{t} = lim_{t\to O^+} \frac{-f(c-tu) + f(c)}{t}. \quad (2).$$
With $(2)$, I have the one side of the limit of $D_{-u} f(c)$, but how to get the other side ?
$$D_uf(c)=\displaystyle\lim_{h\to 0} \frac{f(c+ hu)-f(c)}{h}$$ exists (means it is the same value for either $h<0$ or $h>0$ ) for any unit vector $u$. Look at $$D_{-u}f(c))=\displaystyle\lim_{h\to 0} \frac{f(c-hu)-f(c)}{h}\quad\text{exist}$$ $$=-\displaystyle\lim_{t\to 0} \frac{f(c+tu)-f(c)}{t}=-D_uf(c),$$ where we have used the substitution $t=-h$.