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Let $f$ be a scalar function of three variables. Then the gradient vector is defined by:

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I read here that the existence of partial derivatives at some point $(x_0, y_0, z_0)$ does not imply the existence of the gradient vector at $(x_0, y_0, z_0)$. How is this possible, since the gradient vector is just a specific type of sum of the partial derivatives?

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  • $\begingroup$ First, given $x\in R^{3}$ then $f'(x)$ if it exists is a linear transformation: $\mathbb R^{3}\rightarrow \mathbb R$ which turns out to be $h\mapsto \nabla f(x)\cdot h$. The converse however is not necessarily true. That is, if the partials exist, that does NOT automatically give you differentiability of $f$ at $x$. To see this, look at the counterexamples in you own reference. $\endgroup$ Jun 30 '15 at 23:26
  • $\begingroup$ My question is about the gradient vector, not differentiability. $\endgroup$
    – FreshAir
    Jun 30 '15 at 23:49
  • $\begingroup$ this is semantics: the vector $\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial x}j$ can exist without $f$ being differentiable at $x$. But it is not the gradient, whose name is reserved for the case when $f$ is differentiable. $\endgroup$ Jun 30 '15 at 23:56
  • $\begingroup$ @Chilango thanks for clarifying. But I checked in Stewart's and Edward's texts and did not find this is explicitly stated - ie $\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j$ is only called the gradient when $f$ is differentiable. Would you know of literature that spells this out? $\endgroup$
    – FreshAir
    Jul 1 '15 at 3:03
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Like @Chilango commented, the vector $f_x(x_0, y_0)i+f_y(x_0,y_0)j$ exists even if $f$ is not differentiable at $(x_0, y_0)$.

The link you gave conflates differentiability with existence of the gradient vector, but both my calculus textbook and my real analysis textbook give a different definition for differentiability: A function $f$ is differentiable at $(x_0,y_0)$ if $\Delta z = f_x(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y $ where both $\varepsilon_1$ and $\varepsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$.

The books also show sufficient conditions for differentiability: existence and continuity of both partial derivatives.

Neither book specifies that we only call the vector $f_xi+ f_yj$ the gradient when $f$ is differentiable, but I don't know if this is standard.

Some applications of the gradient do fail if $f_x$ or $f_y$ is not continuous, because they depend on the differentiability of the function. Two examples are the interpretation of the gradient as the direction of maximum increase and the fact that the gradient is normal to the level curves; both of these rely on the existence of the directional derivative. This means that you can't do much with the vector if your function is not differentiable.

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  • $\begingroup$ Thanks for this detailed answer. Actually what you say ($\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial x}j$ is called the gradient regardless of differentiability) seems to be contrary to what @Chilango commented. Also, you say toward the end "both of these rely on the existence of the directional derivative". But don't they rely on more than that fact? Isn't it the case that all directional derivatives can exist at a point, yet the function is not differentiable there? $\endgroup$
    – FreshAir
    Jul 1 '15 at 3:06
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    $\begingroup$ @FreshAir: I guess it's the gradient whether or not $f$ is differentiable. I really never thought about it! You're right that directional derivatives don't imply differentiability either. Take $f(x,y)=\frac{x^{2}y}{x^{4}+y^{2}}$ and $f(0,0)=0$ as an easy counterexample. In fact, $f$ is not even continuous at $(0,0)$. $\endgroup$ Jul 1 '15 at 3:31
  • $\begingroup$ Yes, sorry, not just the existence of the directional derivative, but the fact that it equals the dot product of the gradient and the direction vector, which itself relies on the differentiability of the function. $\endgroup$
    – coldnumber
    Jul 1 '15 at 4:03
  • $\begingroup$ The formula only works if the derivative exists, that's the point. If you compute the directional derivatives from scratch you see that they all exist but the formula $\nabla f(x)\cdot h$ is no good. $\endgroup$ Jul 1 '15 at 4:10

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