# Can or should the limits be one-sided in defintion of directional derivative and proof that it is equal to the differential of the mapping?

In the following, will changing any, or all of the standard limits ( $$\xi\to0$$ ) to one-sided limits ($$\xi\to0^{+}$$) change the meaning of the expressions? Notice in particular, in the proof of Theorem 2.1 $$\xi\left|\Delta\mathfrak{x}\right|$$ is not equivalent to $$\left|\xi\Delta\mathfrak{x}\right|.$$

The definitions and theorem are based on C.H. Edwards's Advanced Calculus of Several Variables, section II-2.

The first difference at $$\mathfrak{x}$$ of the mapping $$\vec{f}:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is defined as $$\Delta\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right)=\vec{f}\left(\mathfrak{x}+\Delta\mathfrak{x}\right)-\vec{f}\left(\mathfrak{x}\right).$$

Assuming the limit exists, the directional derivative with respect to $$\Delta\mathfrak{x}\in\mathbb{R}^{n}$$ of $$\vec{f}:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ at $$\mathfrak{x}$$ is defined to be

\begin{align*} D_{\Delta\mathfrak{x}} & \vec{f}\left(\mathfrak{x}\right)=\lim_{\xi\to0}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)}{\xi}. \end{align*}

The mapping $$\vec{f}:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is said to be differentiable at $$\mathfrak{x}$$ if and only if there exists a linear mapping $$d\vec{f}_{\mathfrak{x}}:\mathbb{R}^{n}\to\mathbb{R}^{m},$$ called the differential of $$\vec{f}$$ at $$\mathfrak{x},$$ such that

$$\mathfrak{0}=\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right)}{\left|\Delta\mathfrak{x}\right|}.$$

Theorem 2.1 If $$\vec{f}:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is differentiable at $$\mathfrak{x},$$ then the directional derivative exists for all $$\Delta\mathfrak{x}\in\mathbb{R}^{n},$$ and

$$D_{\Delta\mathfrak{x}}\vec{f}\left(\mathfrak{x}\right)=d\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right).$$

Proof: In the equation defining differentiability, substitute $$\Delta\mathfrak{x}\mapsto\xi\Delta\mathfrak{x},$$ so that

\begin{align*} \mathfrak{0}= & \lim_{\xi\to0}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)}{\left|\xi\Delta\mathfrak{x}\right|}\\ = & \frac{1}{\left|\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)}{\xi}\right)\\ = & D_{\Delta\mathfrak{x}}\vec{f}\left(\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right). \end{align*}

• Why not define the derivatives at a point $x$ rather than at that weird squiggle thing? You can use $x$ and $y$ and then for a fixed direction you can use $d$ or $v$ and for a scalar to go to zero you can use $h\rightarrow 0$. Jun 23 at 4:19
• Are you talking about the Greek letter $\xi$? feynmanlectures.caltech.edu/II_01.html or perhaps Fraktur font Latin x, $\mathfrak{x}$? johndcook.com/blog/2018/07/21/fraktur-math The reason I used the definition given, is because that's how Edwards does it. Jun 23 at 4:36
• The book Convex Analysis by Rockafellar on page 213 defines the one sided directional derivative of a function $f:\mathbb{R}^n\rightarrow [-\infty, \infty]$ at a point $x$ where $f$ is finite, and with respect to a vector $v$, by $$D_vf(x) = \lim_{h\rightarrow 0^+}\frac{f(x+hv) - f(x)}{h}$$ whenever the limit exists. Actually it uses $f'(x;v)=D_vf(x)$ but I modified to match your notation. It says the directional derivative is "two sided" if $D_{-v}f(x)$ exists and $D_{-v}f(x)=-D_vf(x)$. It says that if $f$ is differentiable at $x$ the directional derivatives are all finite and two-sided. Jun 23 at 4:43

• Rewrite the definition of differentiability using $$\left|\xi\right|$$.
• This is equivalent to writing the limit from the right without the absolute value on $$\xi$$.
• Since the definition applies to all $$\Delta\mathfrak{x},$$ the equation holds if we replace $$\Delta\mathfrak{x}\mapsto{-\Delta\mathfrak{x}}.$$
• Since $$\mathfrak{0}=-\mathfrak{0},$$ the equation holds when we multiply the denominator by $$-1.$$
\begin{align*} \mathfrak{0}= & \lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\Delta\mathfrak{x}\right)}{\left|\Delta\mathfrak{x}\right|}\\ = & \frac{1}{\left|\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\left|\xi\right|\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\left|\xi\right|\Delta\mathfrak{x}\right)}{\left|\xi\right|}\right)\\ = & \frac{1}{\left|\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0^{+}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)}{\xi}\right)\\ = & \frac{1}{\left|-\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0^{+}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(-\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(-\xi\Delta\mathfrak{x}\right)}{\xi}\right)\\ = & -\frac{1}{\left|-\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0^{+}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(-\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(-\xi\Delta\mathfrak{x}\right)}{-\xi}\right)\\ = & -\frac{1}{\left|-\Delta\mathfrak{x}\right|}\left(\lim_{\xi\to0^{-}}\frac{\Delta\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)-d\vec{f}_{\mathfrak{x}}\left(\xi\Delta\mathfrak{x}\right)}{\xi}\right) \end{align*}