Total differentiability implies continuity

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

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

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

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

Then,

\[\lim_{(x,y) \to (a,b)}f(x,y) = f(a,b).\]

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 

\[f(x,y) = \left\{ \begin{array}{cc} \frac{xy}{x^2 + y^2} & ((x,y) \neq (0,0)),\\ 0 & ((x,y) = (0,0)). \end{array} \right.\] 

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

\[\lim_{x\to 0}f(x,mx) = \frac{m}{1 + m^2}.\]

Since this limit clearly depends on \(m\), the limit \(\lim_{(x,y)\to(0,0)}f(x,y)\) 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 \((x,y) \neq (a,b)\). 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

\[f(x,y) - f(a,y) = f_x(h,y)(x - a). \tag{Eq:MVx}\]

Similarly, there exists some \(k\) between \(y\) and \(b\) such that

\[f(a,y) - f(a,b) = f_y(a,k)(y - b).\tag{Eq:MVy}\]

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

\[\begin{eqnarray} \lim_{(x,y)\to (a,b)}f_x(x,y) &=& f_x(a,b),\\ \lim_{(x,y)\to (a,b)}f_y(x,y) &=& f_y(a,b). \end{eqnarray}\]

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

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

as \((x,y) \to (a,b)\). Therefore, 

\[\lim_{(x,y)\to (a,b)}\frac{f(x,y) - f(a,b) - f_x(a,b)(x - a) - f_y(a,b)(y - b)}{\sqrt{(x-a)^2 + (y-b)^2}} = 0,\]

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

\[\begin{eqnarray} f_x(x,y) &=& \frac{g_x(x,y)h(x,y) - g(x,y)h_x(x,y)}{[h(x,y)]^2},\\ f_y(x,y) &=& \frac{g_y(x,y)h(x,y) - g(x,y)h_y(x,y)}{[h(x,y)]^2} \end{eqnarray}\]

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. ■





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