Separately continuous not implies continuous

Statement

For a function of two variables at a point

It is possible to have the following: a function $f$ of two variables $x,y$ that is separately continuous in both variables $x,y$ at the point $(x_0,y_0)$, but is not a continuous function at $(x_0,y_0)$.

For a function of two variables overall

It is possible to have the following: a function $f$ of two variables $x,y$ that is separately continuous in both variables $x,y$ everywhere on ${\mathbb{R}}^2$, but is not continuous everywhere on ${\mathbb{R}}^2$ (i.e., there exist points where it is not continuous.

Related facts

• Existence of partial derivatives not implies differentiable
• Continuous along every linear direction not implies continuous

Proof

We give a single example that illustrates both versions of the statement.

Consider the function:

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

It's clear that $f$ is continuous, as well as separately continuous, at all points other than $(0,0)$. At the point $(0,0)$, we calculate the limits along the $x$-axi:

• The limit along the $x$-direction is $\underset{x\to 0}{lim}\frac{(x)(0)}{x^2+0^2}=\underset{x\to 0}{lim}0=0$. This coincides with the value $f(0,0)$.
• The limit along the $y$-direction is $\underset{y\to 0}{lim}\frac{(0)(y)}{0^2+y^2}=\underset{y\to 0}{lim}0=0$. This coincides with the value $f(0,0)$.

Thus, the function $f$ is separately continuous in both $x$ and $y$ at the point $(0,0)$. Since we already established that it's separately continuous everywhere else, we obtain that it is separately continuous on all of ${\mathbb{R}}^2$.

On the other hand, $f$ is not continuous at $(0,0)$. To see this, consider the limit along the line $y=x$. Setting $y=x$, we get that the limit is:

$\underset{x\to 0}{lim}\frac{xx}{x^2+x^2}=\underset{x\to 0}{lim}\frac{x^2}{2x^2}=\underset{x\to 0}{lim}\frac{1}{2}=\frac{1}{2}\ne f(0,0)$

Note that if the function were indeed continuous at $(0,0)$, the limit along every direction would equal the value at the point, so this shows that the function is not continuous at $(0,0)$.