TITLE: A function continuous in both arguments
QUESTION [5 upvotes]: Is there a two-arguments function which is not continuous but continuous in each argument?
It seems I have studied something like this, but don't remember.

REPLY [9 votes]: Yes, the map $f:\Bbb R^2\to\Bbb R$ $$(x,y)\mapsto\begin{cases}
\dfrac{xy}{x^2+y^2}, &\text{ if }x\ne0\text{ or }y\ne0\\
0, &\text{ if } x=y=0
\end{cases}$$
Then $y\mapsto f(x,y)$ for a fixed $x$ is continuous. As it is symmetric in $x,y$, the map $x\mapsto f(x,y)$ is continuous, too.
But $f$ is not jointly continuous at $(0,0).$ To see this, let $z_n=\left(\frac1n,\frac1n\right)$. Then $f(z_n)=\frac12$ for each $n$. On the other hand, it converges to $(0,0)$ which has image $0.$

REPLY [9 votes]: Consider the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$,
$$
f(x,y) = \begin{cases} 
      \frac{xy}{x^2 + y^2} & (x,y) \neq (0,0) \\
      0 & (x,y) = (0,0) \\
   \end{cases}
$$
This is discontinuous as it is 1/2 arbitrarily close to $(0,0)$ when $x=y$, but it is easily seen to be continuous in each variable.