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It is well known that:

The directional derivative $\nabla_{{v}}{f}$ of a smooth function $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ in the direction of a vector $v$ is defined by: $$\nabla_{{v}}{f}({x}) = \lim_{h \rightarrow 0}{\frac{f({x} + h{v}) - f({x})}{h}}$$. We also know that the gradient of such an $f$ is defined implicitly by the equality $$\lim_{h\to 0} \frac{|f(x+h)-f(x) -\nabla f(x)\cdot h|}{|h|}=0$$.

Can we define (or derive from the latter definition) the gradient as a limit of a difference quotient?

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Assuming $f$ is nice, the gradient at $x$ is the unique vector $\nabla f(x)$ such that for any direction $v$, $$ \nabla f(x)\cdot v = \nabla_v f(x)$$ So by knowing the directional derivatives in $n$ linearly independent directions (e.g. in the direction of the coordinate axes) we can deduce $\nabla f(x)$.

So if you write $F(t)$ for the vector valued function $F(t)_i = f(x+t e_i )$ where $e_i$ are the standard basis vectors, then this allows us to write $$ \nabla f(x) = F'(0) = \lim_{h\to 0} \frac{F(h)-F(0)}h$$

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  • $\begingroup$ An explanation for the down vote would be appreciated $\endgroup$ – Calvin Khor Mar 6 '18 at 8:28
  • $\begingroup$ In your last equation can we replace zero with a minimum point $x^{*}$ of $f$? $\endgroup$ – johnny09 Feb 21 '19 at 1:28
  • $\begingroup$ @johnny09 I have no idea what you mean because the 0 there is a real number, and is the argument of $F$, not $f$ $\endgroup$ – Calvin Khor Feb 21 '19 at 8:48
  • $\begingroup$ The first order optimality condition states that $\nabla f(x^{*})=0$ must be satisfied for $x^{*}$ to be a minimum point. So, I was wondering if one could use your last equation with $x^{*}$. $\endgroup$ – johnny09 Feb 22 '19 at 2:17
  • $\begingroup$ There are no restrictions on $x$ in what I wrote, if that helps? @johnny09 $\endgroup$ – Calvin Khor Feb 22 '19 at 10:01

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