 # Differentiability of multivariable real functions (part1)

This article provides counterexamples about differentiability of functions of several real variables. We focus on real functions of two real variables (defined on $$\mathbb R^2$$). $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. the absolute value for $$\mathbb R$$.

We recall some definitions and theorems about differentiability of functions of several real variables.

Definition 1 We say that a function $$f : \mathbb R^2 \to \mathbb R$$ is differentiable at $$\mathbf{a} \in \mathbb R^2$$ if it exists a (continuous) linear map $$\nabla f(\mathbf{a}) : \mathbb R^2 \to \mathbb R$$ with $\lim\limits_{\mathbf{h} \to 0} \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-\nabla f(\mathbf{a}).\mathbf{h}}{\Vert \mathbf{h} \Vert} = 0$

Definition 2 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. Then the $$\mathbf{i^{th}}$$ partial derivative at point $$\mathbf{a}$$ is the real number
\begin{align*}
\frac{\partial f}{\partial x_i}(\mathbf{a}) &= \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{e_i})- f(\mathbf{a})}{h}\\
&= \lim\limits_{h \to 0} \frac{f(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n) – f(a_1,\dots,a_{i-1},a_i,a_{i+1},\dots,a_n)}{h}
\end{align*} For two real variable functions, $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ will denote the partial derivatives.

Definition 3 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. The directional derivative of $$f$$ along vector $$\mathbf{v}$$ at point $$\mathbf{a}$$ is the real $\nabla_{\mathbf{v}}f(\mathbf{a}) = \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{v})- f(\mathbf{a})}{h}$

Now some theorems about differentiability of functions of several variables.

Theorem 1 Let $$f : \mathbb R^2 \to \mathbb R$$ be a continuous real-valued function. Then $$f$$ is continuously differentiable if and only if the partial derivative functions $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ exist and are continuous.

Theorem 2 Let $$f : \mathbb R^2 \to \mathbb R$$ be differentiable at $$\mathbf{a} \in \mathbb R^2$$. Then the directional derivative exists along any vector $$\mathbf{v}$$, and one has $$\nabla_{\mathbf{v}}f(\mathbf{a}) = \nabla f(\mathbf{a}).\mathbf{v}$$.

### A differentiable function with discontinuous partial derivatives

Consider the function defined on $$\mathbb R^2$$ by
$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & \text{ if } (x,y) \ne (0,0)\\ 0 & \text{ if }(x,y) = (0,0).\end{cases}$ $$f$$ is obviously continuous on $$\mathbb R^2 \setminus \{(0,0)\}$$. $$f$$ is also continuous at $$(0,0)$$ as for $$(x,y) \neq (0,0)$$ $\left\vert (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) \right\vert \le x^2+y^2 = \Vert (x,y) \Vert^2 \mathrel{\mathop{\to}_{(x,y) \to (0,0)}} 0$ $$f$$ is also differentiable at all $$(x,y) \neq (0,0)$$. Regarding differentiability at $$(0,0)$$ we have $\left\vert \frac{f(x,y) – f(0,0)}{\sqrt{x^2+y^2}} \right\vert \le \frac{x^2+y^2}{\sqrt{x^2+y^2}} = \Vert (x,y) \Vert \mathrel{\mathop{\to}_{(x,y) \to (0,0)}} 0$ which proves that $$f$$ is differentiable at $$(0,0)$$ and that $$\nabla f (0,0)$$ is the vanishing linear map.

The partial derivatives of $$f$$ are zero at the origin. For example, the derivative with respect to $$x$$ can be calculated by
\begin{align*}
\frac{\partial f}{\partial x}(0,0) &= \lim_{h \to 0}
\frac{f(h,0)-f(0,0)}{h}\\
&= \lim_{h \to 0}\frac{h^2 \sin (1/|h|)-0}{h} \\
&= \lim_{h \to 0}h \sin (1/|h|) =0.
\end{align*}
A similar calculation shows that $$\frac{\partial f}{\partial x}(0,0)=0$$.

Away from the origin, one can use the standard differentiation formulas to calculate that
\begin{align*}
\frac{\partial f}{\partial x}(x,y) &= 2 x \sin
\left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{x \cos
\left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}\\
\frac{\partial f}{\partial y}(x,y) &= 2 y \sin
\left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{y \cos
\left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}.
\end{align*}
Both of these derivatives oscillate wildly near the origin. For example, the derivative with respect to $$x$$ along the $$x$$-axis is $$\frac{\partial f}{\partial x}(x,0) = 2 x \sin \left(1/|x|\right)-\text{sign}(x) \cos \left(1/|x|\right),$$
for $$x \neq 0$$, where $$\text{sign}(x)$$ is $$\pm 1$$ depending on the sign of $$x$$. In this case, the sine term goes to zero near the origin but the cosine term oscillates rapidly between $$-1$$ and $$+1$$. Hence $$\frac{\partial f}{\partial x}$$ is discontinuous at the origin. In the same way, one can show that $$\frac{\partial f}{\partial y}$$ is discontinuous at the origin.

This counterexample proves that theorem 1 cannot be applied to a differentiable function in order to assert the existence of the partial derivatives. Continuity of the derivative is absolutely required!

### A function having partial derivatives which is not differentiable

We now consider the converse case and look at $$g$$ defined by
$g(x,y)=\begin{cases}\frac{xy}{\sqrt{x^2+y^2}} & \text{ if } (x,y) \ne (0,0)\\ 0 & \text{ if }(x,y) = (0,0).\end{cases}$ For all $$(x,y) \in \mathbb R^2$$ we have $$x^2 \le x^2+y^2$$ hence $$\vert x \vert \le \sqrt{x^2+y^2}=\Vert (x,y) \Vert$$. Similarly, $$\vert y \vert \le \Vert (x,y) \Vert$$ and therefore $$\vert g(x,y) \vert \le \Vert (x,y) \Vert$$. This last inequality being also valid at the origin. Consequently, $$g$$ is a continuous function. The partial maps $$x \mapsto g(x,0)$$ and $$y \mapsto g(0,y)$$ are always vanishing. Hence $$g$$ has partial derivatives equal to zero at the origin.

As a consequence, if $$g$$ was differentiable at the origin, its derivative would be equal to zero and we would have $\lim\limits_{(x,y) \to (0,0)} \frac{\vert g(x,y) \vert}{\Vert (x,y) \Vert} = 0$ That is not the case as for $$x \neq 0$$ we have $$\displaystyle \frac{\vert g(x,y) \vert}{\Vert (x,y) \Vert} = \frac{1}{2}$$. Finally $$f$$ is not differentiable.

We also have $\frac{\partial g}{\partial x}(x,y) = \frac{y^3}{(x^2+y^2)^{\frac{3}{2}}}, \frac{\partial g}{\partial x}(0,y) = \text{sign}(y)$ which proves that $$\frac{\partial g}{\partial x}$$ is not continuous at the origin avoiding any contradiction with theorem 1.

### A function having directional derivatives along all directions which is not differentiable

We prove that $$h$$ defined by
$h(x,y)=\begin{cases}\frac{x^2 y}{x^6+y^2} & \text{ if } (x,y) \ne (0,0)\\ 0 & \text{ if }(x,y) = (0,0)\end{cases}$ has directional derivatives along all directions at the origin, but is not differentiable at the origin.

First of all, $$h$$ is a rational fraction whose denominator is not vanishing for $$(x,y) \neq (0,0)$$. Hence $$h$$ is continuously differentiable for $$(x,y) \neq (0,0)$$. Let’s have a look to the directional derivatives at the origin. Let’s fix $$\mathbf{v} = (\cos \theta, \sin \theta)$$ with $$\theta \in [0, 2\pi)$$. For $$t \neq 0$$, we have $\frac{h(t \cos \theta, t \sin \theta) – h(0,0)}{t}= \frac{ \cos^2 \theta \sin \theta}{t^4\cos^6 \theta + \sin^2 \theta}$ which goes to $$0$$ as $$t \to 0$$ whatever the value of $$\theta$$. Therefore, $$h$$ has directional derivatives along all directions at the origin.

However, $$h$$ is not differentiable at the origin. In fact $$h$$ is not even continuous at the origin as we have $h(x,x^3) = \frac{x^2 x^3}{x^6 + (x^3)^2} = \frac{1}{x}$ for $$x \neq 0$$.