2
$\begingroup$

I am reading Burkill's "A Second Course in Mathematical Analysis". (On page 198) it says that

the existence of all directional derivatives at a point does not ensure continuity at that point.

But then on page 201, it says

If $X\subset \mathbb R^m$ and the function $f:X\to\mathbb R^n$ is differentiable at the interior point $\xi$ of $X$, then $f$ is continuous at $\xi$.

Can someone explain how these do not contradict? I am guessing the key pertains to the "existence of all directional derivatives" v.s. "differentiable"? Grateful if someone could point out the distinction.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ Look for definition of 'directional derivatives' and of 'being differentiable'. They look similar but there is a great difference. If a function is differentiable at a point, there exists all directional derivates, but not the convese. $\endgroup$ – Rafael Mrđen Nov 21 '11 at 11:44
1
$\begingroup$

Your guess is correct; given $X\subseteq \mathbb{R}^m$ and a function $f:X\to \mathbb{R}^n$, then $f$ is differentiable at $a\in\mathbb{R}^n$ when there is a linear transformation $L:\mathbb{R}^n\to\mathbb{R}^m$ such that $$\lim_{h\to 0}\frac{|f(a+h)-f(a)-L(h)|}{|h|}=0,$$ and when it exists, this $L$ is called $Df(a)$, the derivative of $f$ at $a$, and it is unique. If $m=1$ (i.e. the codomain of $f$ is $\mathbb{R}$), then the directional derivative of $f$ at $a\in\mathbb{R}^n$ in the direction of $x\in\mathbb{R}^n$ is $$\lim_{t\to 0}\frac{f(a+xt)-f(a)}{t}.$$ The differentiability of $f$ at $a$ implies the existence of all directional derivatives at $a$, but not conversely. However, if one assumes that there is an open set $U\subseteq \mathbb{R}^n$ containing $a$ such that all directional derivatives of $f$ exist everywhere in $U$, and the directional derivatives are continuous at $a$, then we are guaranteed that $Df(a)$ exists (this is Theorem 2-8 in Spivak's Calculus on Manifolds).

Thus, if we want a counterexample, we should look for functions whose directional derivatives are either not defined, or not continuous, at or around the point $a$. The classic example is the characteristic function of $$A=\{(x,y)\in\mathbb{R}^2\mid x>0,\;\;0<y<x^2\},$$ i.e. the function $f$ that is defined by $f(x)=1$ if $x\in A$ and $f(x)=0$ if $x\notin A$. Intuitively, one can see that all directional derivatives of $f$ exist at the origin $(0,0)$, because in any straight direction from the origin, there is some initial distance one can travel in which one is not in $A$, so all the directional derivatives of $f$ exist at $(0,0)$ and are equal to $0$. However, the limit defining the derivative $Df(0,0)$ "can see all paths" to the origin, including those that remain contained inside $A$ all the way there. Even though all the directional derivatives of $f$ exist and are continuous at $(0,0)$, the directional derivatives of $f$ do not exist at any other points along the boundary of $A$, and any neighborhood of the origin must contain some of those boundary points.

Note that, even if it were true that the existence of all partial derivatives at a point guaranteed the existence of the derivative at that point, the set of points where all the partial derivatives of $f$ exists is $A\cup \text{int}(A^c)\cup\{(0,0)\}$, and $(0,0)$ is not an interior point of that set.

$\endgroup$
  • $\begingroup$ No problem, glad to help! $\endgroup$ – Zev Chonoles Nov 21 '11 at 12:30

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.