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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*}

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  • $\begingroup$ 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 $h\rightarrow 0$. $\endgroup$
    – Michael
    Jun 23 at 4:19
  • $\begingroup$ 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. $\endgroup$ Jun 23 at 4:36
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Jun 23 at 4:43

1 Answer 1

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As pointed out in the comments, making the limit in the definition of the directional derivative one-sided changes the meaning of the definition. It may be instructive or useful to examine such one-sided limits, but the standard limit provides the common definition of the directional derivative.

Regarding the limit appearing in the proof of Theorem 2.1, the following seems to justify using a standard (two-sided) limit.

  • 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 $-1.$
  • This produces the limit from the left of the same expression for which we originally wrote the limit from the right.
  • Thus we have a two-sided limit.

\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*}

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