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Given a function $f(x,y)$, separate continuity means fixing $y=y_0$ then $f(x,y_0)$ is continuous with respect with $x$, and fixing $x=x_0$ then $f(x_0,y)$ is continuous with respect with $y$. Joint continuity simply means $f(x,y)$ is continuous.

It has been pointed out in other posts that separate continuity does not imply joint continuity, for example $$f(x,y) = \begin{cases}0, & x = y = 0\\[2ex] \dfrac{xy}{x^2+y^2}, & x^2+y^2 > 0\end{cases}$$ has separate continuity on $[-1,1]$ but does not have joint continuity.

So the following reasoning should be incorrect, but where am I going wrong with the following reasoning? Can anyone help explain? Thank you!

Suppose $f(x,y)$ has separate continuity. For any $(x,y)$ and any sequence $(x_n,y_m)\rightarrow (x,y)$, first fix $m$ and we have $f(x_n,y_m)\rightarrow f(x,y_m)$, then we have $f(x,y_m)\rightarrow f(x,y)$. Thus $f(x_n,y_m)\rightarrow f(x,y)$ for any $(x,y)$ and $(x_n,y_m)\rightarrow (x,y)$ and hence $f(x,y)$ is continuous.

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Tony
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1 Answers1

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The problem with this is that $f(x_n,y_m)\to f(x,y_m)$ and $f(x,y_m)\to f(x,y)$ does not imply that $f(x_n,y_m)\to f(x,y)$. If you try to prove this, you'll find that you actually need that $f(x_n,y_m)$ converges to $f(x,y_m)$ uniformly in $m$. That is, you need to know that for any $\epsilon>0$, there is an $N$ such that $|f(x_n,y_m)-f(x,y_m)|<\epsilon$ for all $n>N$ and all $m$, with $N$ not depending on $m$. There is no reason for this to be true in general. It is instructive to look at what happens with the counterexample function $f$ you describe and the sequence $(x_n,y_m)=(1/n,1/m)$.

Eric Wofsey
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