So I've been asked to prove non-differentiability using this particular method, and I'm a bit lost. I'm supposed to prove that $f(x,y) = |x-y|$ is not differentiable along the $x=y$ line. To do it, I'm supposed to use this definition of differentiability:

A function $f$ is called differentiable at a point $a$ in it's domain if there exists a vector $c\in\mathbb{R^{n}}$ such that:

$$\lim\limits_{\bf h \to \bf 0} \frac{f(\boldsymbol a + \boldsymbol h) - f(\boldsymbol a) - \boldsymbol c \cdot \boldsymbol h}{|h|} = 0$$

Okay, so obviously $c$ would be the gradient of $f$ at that point, and obviously it won't exist because there is an apex there, but I can't for the life of me prove that it doesn't exist. Any help would be appreciated.

  • $\begingroup$ Do you know that if a function is differentiable at a point, then all directional derivatives exist at that point? $\endgroup$ – wj32 Nov 22 '12 at 4:24
  • $\begingroup$ yeah I do... I guess that follows from this definition, since c is grad(f)... So I guess I could use that fact. Its annoying that they're asking to use this method... $\endgroup$ – Harris M Snyder Nov 22 '12 at 5:11

Consider the point ${\bf a}=(a,a)$ for some $a\in{\mathbb R}$. Assume that $f$ is differentiable at ${\bf a}$ according to the above definition. As $f({\bf a})=0$ we would then simultaneously have $$\lim_{{\bf h}\to{\bf 0}}{f({\bf a}+{\bf h})-{\bf c}\cdot {\bf h}\over |{\bf h}|}=0\ ,\quad \lim_{{\bf h}\to{\bf 0}}{f({\bf a}-{\bf h})-{\bf c}\cdot(-{\bf h})\over |{\bf h}|}=0\ ,$$ and therefore $$\lim_{{\bf h}\to{\bf 0}}{f({\bf a}+{\bf h})+f({\bf a}-{\bf h})\over |{\bf h}|}=0\ .\qquad(*)$$ For real $h>0$ put ${\bf h}:=(h,-h)$. Then $|{\bf h}|=\sqrt{2} h$ and $$f({\bf a}+{\bf h})+f({\bf a}-{\bf h})=\bigl|(a+h)-(a-h)\bigr|+\bigl|(a-h)-(a+h)\bigr|=4h\ .$$ As $$\lim_{h\to 0+}{4h \over\sqrt{2} h}=2\sqrt{2}\ne0\ ,$$ condition $(*)$ is not fulfilled at this point ${\bf a}$.

  • $\begingroup$ Okay, I follow that logic: a + h and a - h are approaching the apex from opposite directions. I'm not sure that I understand why that equation follows from the definition though. Basically negative c dotted with h has to equal f(a-h) - f(a), which I only kind of understand. Are we just saying that the limit must be the same when approached from above and below? $\endgroup$ – Harris M Snyder Nov 22 '12 at 21:37
  • $\begingroup$ @Harris M Snyder: See my edit; it should be clear now. $\endgroup$ – Christian Blatter Nov 23 '12 at 9:22
  • $\begingroup$ Ah yes now I see, thank you very much $\endgroup$ – Harris M Snyder Nov 25 '12 at 17:33

Let $x \in \mathbb{R}$, $h>0$, then

$$\left|\frac{f((x,x)^T+h \cdot e^1)-f((x,x)^T)-1 \cdot h}{|h|} \right| = \left| \frac{|(h,0)^T|-1 \cdot h}{h} \right| \stackrel{h>0}{=} 0$$

On the other hand

$$\left|\frac{f((x,x)^T+h \cdot e^1)-f((x,x)^T)+1 \cdot h}{|h|} \right| = \left| \frac{|(h,0)^T|+1 \cdot h}{-h} \right| \stackrel{h<0}{=} 0$$

for $h<0$ where $e^1:=(1,0)^T$. This means that $f$ is not differentiable at $(x,x)$ (...since the previous equations show that $\partial_x f$ does not exist. Or if you want to prove it straight from the definition of differentiability: The first equation equation shows $c \cdot (h \cdot e^1) =c_1 \cdot h \stackrel{!}{=} 1 \cdot h$ whereas the second one shows $c \cdot (h \cdot e^1) = c_1 \cdot h \stackrel{!}{=} -1 \cdot h$ - and there exists obviously no such $c=(c_1,c_2) \in \mathbb{R}^2$.)

Similar proof works for $n \geq 3$.

  • $\begingroup$ I'm not completely able to follow. Are you using superscript T to represent matrix transpose? $\endgroup$ – Harris M Snyder Nov 22 '12 at 23:04
  • $\begingroup$ Yes, that's right. If you prefer it, you can simply forget about the $T$ and read $(\cdot,\cdot)$ as a vector in $\mathbb{R}^2$. $\endgroup$ – saz Nov 23 '12 at 6:54
  • $\begingroup$ Thank you, this post was helpful $\endgroup$ – Harris M Snyder Nov 25 '12 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.