Total differentiability implies continuity

Recall that the function $f(x,y)$ is totally differentiable at $(a,b)$ if 

$$f(x,y)=f(a,b)+m(x-a)+n(y-b)+o(\Vert X-P\Vert )\text{ as }X=(x,y)\to P=(a,b)$$

for some constants $n$ and $m$.

See also: Partial and total differentiation of multivariate functions.

Just like univariate functions, we have the following result:

Theorem (Total differentiability implies continuity)

If the function $f(x,y)$ is totally differentiable at $(a,b)$, then it is continuous at $(a,b)$.

Proof. Let us use the notations $X=(x,y)$ and $P=(a,b)$. By the definition of total differentiability, we have

$$f(x,y)=f(a,b)+m(x-a)+n(y-b)+o(\Vert X-P\Vert )$$ for some constants $n,m$.

Then,

$$\underset{(x,y)\to (a,b)}{lim}f(x,y)=f(a,b).$$

Hence, $f(x,y)$ is continuous at $(a,b)$. ■

Remark. Just as for the case of univariate functions, the converse of this theorem does not hold. For example, you should verify that $f(x,y)=|x-y|$ is continuous everywhere in ${\mathbb{R}}^2$, but not totally differentiable on the line $y=x$. □

Example. Consider the function 

$$f(x,y)=\{\begin{array}{cc}\frac{xy}{x^2+y^2}& ((x,y)\ne (0,0)),\\ 0& ((x,y)=(0,0)).\end{array}$$ 

Let $y=mx$ and consider the limit $f(x,y)$ as $x\to 0$:

$$\underset{x\to 0}{lim}f(x,mx)=\frac{m}{1+m^2}.$$

Since this limit clearly depends on $m$, the limit $\underset{(x,y)\to (0,0)}{lim}f(x,y)$ does not exist, and hence $f(x,y)$ is not continuous at $(0,0)$. By the contrapositive of the above theorem, $f(x,y)$ is not totally differentiable.

Nevertheless, since $f(x,0)=0$, $f_x(0,0)=0$; and since $f(0,y)=0$, $f_y(0,0)=0$. Thus, even though $f(x,y)$ is not totally differentiable at $(0,0)$, its partial differential coefficients exist at $(0,0)$. □

As this example shows, the existence of partial differential coefficients does not guarantee total differentiability. However, we have the following result.

Theorem (Criterion of total differentiability)

Let $U$ be an open region in ${\mathbb{R}}^2$ and $(a,b)\in U$. Let $f(x,y)$ be a function on $U$. If the partial derivatives $f_x(x,y)$ and $f_y(x,y)$ exist on $U$ and are continuous at $(a,b)$, then $f(x,y)$ is totally differentiable at $(a,b)$.

Proof. Let $(x,y)\in U$ be an arbitrary point such that $(x,y)\ne (a,b)$. Since $f_x(x,y)$ exists on $U$, $f(x,y)$ is continuous as a function of $x$. Thus, by the Mean Value Theorem (for univariate functions), there exists some $h$ between $x$ and $a$ such that

$$\begin{array}{}\text{(Eq:MVx)}& f(x,y)-f(a,y)=f_x(h,y)(x-a).\end{array}$$

Similarly, there exists some $k$ between $y$ and $b$ such that

$$\begin{array}{}\text{(Eq:MVy)}& f(a,y)-f(a,b)=f_y(a,k)(y-b).\end{array}$$

By assumption, $f_x(x,y)$ and $f_y(x,y)$ are continuous at $(a,b)$ so that

$$\begin{array}{rcl}\underset{(x,y)\to (a,b)}{lim}{f}_x(x,y)& =& f_x(a,b),\\ \underset{(x,y)\to (a,b)}{lim}{f}_y(x,y)& =& f_y(a,b).\end{array}$$

Let $s=x-a$ and $t=y-b$. From Eqs. (Eq:MVx) and (Eq:MVy) above, and noting 

$$f(x,y)-f(a,b)=f(x,y)-f(a,y)+f(a,y)-f(a,b),$$

we have

$$\begin{array}{rcl}& & \left|\frac{f(x,y)-f(a,b)-f_x(a,b)(x-a)-f_y(a,b)(y-b)}{\sqrt{(x-a{)}^2+(y-b{)}^2}}\right|\\ & & =\left|\frac{\{f_x(h,y)-f_x(a,b)\}s+\{f_y(a,k)-f_y(a,b)\}t}{\sqrt{s^2+t^2}}\right|\\ & & \le |f_x(h,y)-f_x(a,b)|\frac{|s|}{\sqrt{s^2+t^2}}+|f_y(a,k)-f_y(a,b)|\frac{|t|}{\sqrt{s^2+t^2}}\\ & & \le |f_x(h,y)-f_x(a,b)|+|f_y(a,k)-f_y(a,b)|\to 0\end{array}$$

as $(x,y)\to (a,b)$. Therefore, 

$$\underset{(x,y)\to (a,b)}{lim}\frac{f(x,y)-f(a,b)-f_x(a,b)(x-a)-f_y(a,b)(y-b)}{\sqrt{(x-a{)}^2+(y-b{)}^2}}=0,$$

which shows that $f(x,y)$ is totally differentiable at $(a,b)$. ■

Example. With polynomial functions $g(x,y)$ and $h(x,y)$, the rational function given by

$$f(x,y)=\frac{g(x,y)}{h(x,y)}$$

is continuous everywhere except where $h(x,y)=0$. $f(x,y)$ has the partial derivatives

$$\begin{array}{rcl}{f}_x(x,y)& =& \frac{g_x(x,y)h(x,y)-g(x,y)h_x(x,y)}{[h(x,y){]}^2},\\ f_y(x,y)& =& \frac{g_y(x,y)h(x,y)-g(x,y)h_y(x,y)}{[h(x,y){]}^2}\end{array}$$

which are also rational functions of $x$ and $y$, and hence are continuous everywhere except where $h(x,y)=0$. By the above theorem, $f(x,y)$ is totally differentiable everywhere except where $h(x,y)=0$. □

Definition (Continuously differentiable, $C^1$)

The function $f(x,y)$ is said to be continuously differentiable or of class $C^1$ function if the partial derivatives $f_x(x,y)$ and $f_y(x,y)$ exist and are continuous.

Corollary. 

A continuously differentiable function is totally differentiable and continuous.

Proof. Exercise. ■