# Partial Derivatives Continuous does not guarantee Gradient Function continuous

Just want to check my understanding that partial derivatives continuous does not mean that the gradient function $\nabla f$ is continuous. Is that correct?

E.g. $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ continuous does not necessarily mean that $\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ is continuous.

I concluded that based on "Continuity in each argument is not sufficient for multivariate continuity".

Question 2) What conditions guarantee that $\nabla f$ is continuous?

Thanks for any help.

We have that $\nabla f$ is continuous. Note that:

\begin{align*} || \nabla f(x,y) - \nabla f(x_0,y_0)||^2 = (\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0))^2 + (\frac{\partial f}{\partial y}(x,y)- \frac{\partial f}{\partial y}(x_0,y_0))^2 \end{align*} Thus continuity of the partial derivatives imply that the gradient is continuous.

Your quote says something different too. It says that continuity in x and y resp. does not imply continuity in $(x,y)\in \mathbb R^2$. This makes perfect sence, in $\mathbb R^2$ you can approach a point from many more directions that just parallel along the $x$- and $y$-axis.

DETAILS: Let $\epsilon >0$ be fixed. Then by continuity of the partial derivatives we know there exist $\delta_1 >0$ and $\delta_2>0$ so that:

\begin{align*} ||(x,y)-(x_0,y_0)|| <\delta_1 \quad \Rightarrow \quad|\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0)| < \frac{\epsilon}{\sqrt{8}}\\ ||(x,y)-(x_0,y_0)|| <\delta_2 \quad \Rightarrow \quad|\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0)| < \frac{\epsilon}{\sqrt{8}} \end{align*}

Now let $\delta = \min \{\delta_1, \delta_2 \}$. Now we have that:

\begin{align*} || \nabla f(x,y) - \nabla f(x_0,y_0)||^2 &= (\frac{\partial f}{\partial x}(x,y)- \frac{\partial f}{\partial x}(x_0,y_0))^2 + (\frac{\partial f}{\partial y}(x,y)- \frac{\partial f}{\partial y}(x_0,y_0))^2\\ & < \frac{2\epsilon^2}{8} = \frac{\epsilon^2}{4}. \end{align*} Thus we have: \begin{align*} || \nabla f(x,y) - \nabla f(x_0,y_0)|| < \frac{\epsilon}{2}, \end{align*} as required.

Any vector-valued function ${\bf g}=(g_1,\ldots,g_m)$ is continuous (at some point $p$ or overall) iff all its component functions $g_i$ are continuous. This is an immediate consequence of the following inequalities for vectors ${\bf x}$, ${\bf y}$ in ${\mathbb R}^n$: $$|x_i-y_i|\leq|{\bf x}-{\bf y}|\leq\sum_{k=1}^n|x_k-y_k|\ .$$ In particular ${\bf g}:=\nabla f$ is continuous iff all coordinates ${\partial f\over\partial x_k}$ of $\nabla f$ are continuous.

One way things can go wrong is if the partial derivatives exist but the gradient does not. For instance, $$f(x,y) = \left\{ \begin{array}{cl} \dfrac{xy}{x^2 + y^2} & \text{if }(x,y) \not= (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{array} \right.$$ has the property that $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ exist and are continuous at all points $(x,y)$, but $f$ has no gradient at the origin as it fails even to be continuous there.

It is for this reason that writing $\nabla f = ( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$ requires a small amount of caution.

• But if all partial derivatives exist and are continuous everywhere doesn't it follow that f is differentiable in any point and therefore also continuous? – firstsnow Jun 20 '18 at 19:34