Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let us assume that directional derivative of a function $f$ exists at a point $p$ (i.e.,$ D_v(f)$) for all vectors $v \in \mathbb{R}^{n}$. Does it imply that the function is differentiable?

share|cite|improve this question
Nope. See for instance. –  Akhil Mathew Mar 11 '11 at 3:24
You need all directional derivative to be continuous or similar conditions to prove that. –  Kerry Mar 11 '11 at 4:50
@Subramani: Please see Kumaresan's notes on A Conceptual Introduction to Multivariable Calculus. –  anonymous Mar 12 '11 at 6:25

2 Answers 2

Subramani: Let $f\colon U\rightarrow V$ be defined where $U\subset M$ and $V\subset N,$ and $M,N$ are two manifolds. Every manifold comes with tangent spaces. If $x\in M$ then the tangent space at $x$ is the space of "directional derivative" operators. Indeed, locally for any nice curve passing though $x$ one can differentiate $f$ at $x$ in the direction of that curve, which in return, gives a quantity which describes the rate of change of $f$ on this curve at $x.$ The space of maps $D_{\gamma}$ ,whose input is the function and whose output is the directional derivative of that function, is the tangent space.

Differentiability is a different thing: Namely, $f$ is differentiable at $x\in U$ if (1) There is a linear map between the tangent space at $x$ and the tangent space of $f(x)$ (2) This maps approximates $f$ near the point $x.$

When (1) and (2) both holds one can compute $f$ locally near $x$ via the linear map. But the domain and range of this linear map are linear spaces, not the points of manifolds. To approximate $f,$ one projects the tangent space onto the manifold. In summary:

(*) a function can have directional derivative along only one curve, but to define the derivative it should have directional derivatives in all directions+ they should form a linear map

share|cite|improve this answer

You need to recall the definition of differentiability for functions on $\mathbb{R}^n$. The standard definition is

Definition A function $f$ defined on $\mathbb{R}^n$ taking values in $\mathbb{R}$ (or $\mathbb{C}$) is said to be differentiable at $p\in\mathbb{R}^n$ if there exists a linear map $J_p:\mathbb{R}^n\to\mathbb{R}$ (or $\mathbb{C}$) such that for all vectors $v$ in $\mathbb{R}^n$ you have that $$ \lim_{s\to 0^+} \frac{ f(p + sv) - f(p) - J_p(sv)}{s} = 0$$

The key thing to note is that the mapping induced by the "directional derivatives" that sends a vector $v$ to the derivative $D_v f$ is required to be linear map! (This reflects our intuitive understanding that a derivative gives a linear approximation to the function near a point.) So in your situation, if you can prove that the map $v \mapsto D_v f$ is linear, that is, given any $v,w\in\mathbb{R}^n$ and $a,b\in\mathbb{R}$ you have

$$ av + bw \mapsto a D_v f + bD_w f $$

then you have that the function is differentiable. The example linked to by Akhil in his comment shows an example where this linearity fails, and hence the function, depsite having all directional derivatives, is not differentiable.

share|cite|improve this answer
Wow, thanks for this answer Willie, this also shows why we would want the covariant derivative to be bilinear, I guess (I hadn't realized this until now). –  Sam Mar 11 '11 at 12:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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