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Q. Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$. For each vector $\mathbf{v} \in \mathbb{R}^{n}$, we define $$ D_{\mathbf{v}} f(\mathbf{a})=\lim _{t \rightarrow 0} \frac{f(\mathbf{a}+t \mathbf{v})-f(\mathbf{a})}{t} $$ if the limit exists. $D_{\mathbf{v}} f(\mathbf{a})$ is the directional derivative of $f$ with respect to $v$ at $a$. Show that for vectors $\mathbf{v}, \mathbf{w} \in \mathbb{R}^{n},$ one has $$ D_{\mathbf{v}+\mathbf{w}} f(\mathbf{a})=D_{\mathbf{v}} f(\mathbf{a})+D_{\mathbf{w}} f(\mathbf{a}) $$

My attempt: $$ \begin{array}{l}\lim _{t \rightarrow 0} \frac{f(a+t(\mathbf v+\mathbf w))-f(a)}{t} \\ =\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a)}{t}\\ =\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a+t\mathbf v)}{t}+\frac{ f(a+t\mathbf v)-f(a)}{t} \end{array} $$ Now, I have to prove that $$\operatorname{lim}_{t \rightarrow 0}\frac{ f(a+t\mathbf v+t\mathbf w)-f(a+t\mathbf v)}{t}=D_{\mathbf{w}} f(\mathbf{a}) $$ But how to?

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    $\begingroup$ Can we assume that the directional derivative exists for all $v\in\mathbb{R}^n$? $\endgroup$
    – Philipp
    Jul 28, 2020 at 20:07

5 Answers 5

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The counter-example is made of pieces of a cone:

$$z=f(x,y)=\text{sgn}(x)\sqrt{|xy|}.$$

Clearly $f(x,0)=f(0,y)=0$ for all $x$ or $y$, so the directional derivatives along $\mathbf v=(1,0)$ and $\mathbf w=(0,1)$ are both $0$.

But the directional derivative along $\mathbf v+\mathbf w=(1,1)$ is

$$\lim_{t\to0}\frac{f(t,t)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(t)\sqrt{t^2}}{t}$$

$$=\lim_{t\to0}\frac{\text{sgn}(t)\,|t|}{t}=\lim_{t\to0}\frac{t}{t}=1.$$

More generally, the directional derivative along $(a,b)$ is

$$\lim_{t\to0}\frac{f(at,bt)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(a)\text{sgn}(t)\sqrt{|ab|}\,|t|}{t}$$

$$=\text{sgn}(a)\sqrt{|ab|}.$$

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  • $\begingroup$ See the graph here. $\endgroup$
    – mr_e_man
    Jul 28, 2020 at 20:49
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    $\begingroup$ Very nice counterexample! And you can even simplify it more, I believe, taking simply $\;f(x,y)=\sqrt{|xy|}\;$ and, again, taking the dir. deriv. on the same directions on the origin. $\endgroup$
    – DonAntonio
    Jul 28, 2020 at 20:53
  • $\begingroup$ No, that doesn't work with two sided derivatives; the limit with $t\to0^+$ is $1$, but the limit with $t\to0^-$ is $-1$. $\endgroup$
    – mr_e_man
    Jul 28, 2020 at 20:54
  • $\begingroup$ @mr I don't understand that: it anyways fulfills $\;f(0,y)=f(x,0)=f(0,0)\;$, and thus both dir. der. at the origin in the directions $\;(1,0),\,(0,1)\;$ are zero, but not so in the direction of $\;(1,1)\;$ since the limit doesn't even exist... $\endgroup$
    – DonAntonio
    Jul 28, 2020 at 20:57
  • $\begingroup$ Okay, I guess I was assuming that the limit should exist. See Philipp's comment to the OP. $\endgroup$
    – mr_e_man
    Jul 28, 2020 at 20:58
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If $\;f\;$ is differentiable at $\;a\;$ it is pretty easy, since then by linearity of the scalar product (or inner product, to name it with other words) we get

$$D_{v+w}f(a)=\nabla f(a)\cdot(v+w)=\nabla f(a)\cdot v+\nabla f(a)\cdot w=D_vf(a)+D_wf(a)$$

If $\;f\;$ isn't differentiable then it may not be true...but I can't produce a counterexample right now.

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Assuming that $f$ is differentiable at $x=a$ we have

$$f(\mathbf a+t\mathbf v+t\mathbf w)=f(\mathbf a)+\nabla f(\mathbf a)\cdot(t\mathbf v+t\mathbf w)+o(|t\mathbf v+t\mathbf w|)=\\=f(\mathbf a)+\nabla f(\mathbf a)\cdot t\mathbf v+\nabla f(a)\cdot t\mathbf w+o(|t\mathbf v+t\mathbf w|)$$

therefore

$$\frac{ f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf a)}t=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w+\frac{o(|t\mathbf v+t\mathbf w|)}{t}$$

and

$$\lim _{t \rightarrow 0} \frac{f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf{a})}{t}=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w=D_{\mathbf{v}} f(\mathbf{a})+D_{\mathbf{w}} f(\mathbf{a})$$

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  • $\begingroup$ Yes. in that case it is pretty easy (and in fact the proof is way shorter...), but what if $\;f\;$ isn't differentiable...? $\endgroup$
    – DonAntonio
    Jul 28, 2020 at 20:16
  • $\begingroup$ @DonAntonio I suppose it is not true if f is not diffrentiable, but I don't have a proof! Can we find a couterexample. $\endgroup$
    – user
    Jul 28, 2020 at 20:18
  • $\begingroup$ @user That's what I suspect also... $\endgroup$
    – DonAntonio
    Jul 28, 2020 at 20:19
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As a counter example when $f$ is not differentiable, suppose $f(0) = 0$. Then take $f(x) = 0$ along the lines $\{ te_1 \in \mathbb{R}^n \ | \ t\in\mathbb{R}\}$ and $\{t e_2\in \mathbb{R}^n \ | \ t\in\mathbb{R} \}$ then $$D_{e_1}f(0) = 0, \ D_{e_2}f(0) = 0$$ Set $f(x) = \text{sgn}(x_1)|x|$ for $x\in \{ t(e_1+e_2)\in \mathbb{R}^n \ | \ t\in\mathbb{R} \}$ then \begin{align} D_{e_1+e_2}f(0) &= \lim_{t\rightarrow 0}\frac{f(t(e_1+e_2)) - f(0)}{t} \\ &= \lim_{t\rightarrow 0}\frac{\sqrt{2}|t|\text{sgn}(t) - 0}{t} \\ &= \sqrt{2} \\ &\neq D_{e_1}f(0) + D_{e_2}f(0) \end{align}

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Let $f:\mathbb{R}^2 \to \mathbb{R}$ be one on the axes and zero everywhere else.

Then $D_{e_k} f(0) = 0$ but the directional derivative in any other direction does not exist.

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