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Consider a function $f(x,y):[0,1] \times [0,1] \rightarrow R.$ What is the difference between $f$ continuous in each argument and jointly continuous?

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the difference is in definitions, so you may want to find an example what the function is continuous in each argument but not jointly – Ilya Jan 13 '12 at 17:58
Also see an essentially same question Does measurability/continuity of a mapping follow that of its sections? – Tim Jan 14 '12 at 14:53
up vote 8 down vote accepted

Consider the function $f:[0,1]^2\to\mathbb R$ defined by $f(x,y)=\frac{2xy}{x^2+y^2}$ if $(x,y)\ne(0,0)$ and $f(0,0)=0$. Then $f$ is discontinuous at $(0,0)$ (and continuous everywhere else) because $f(t,t)=1$ for every $t\ne0$. But both functions $f(0,\ ):t\mapsto f(0,t)$ and $f(\ ,0):t\mapsto f(t,0)$ are continuous everywhere since $f(0,t)=f(t,0)=0$ for every $t$.

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+1 Great minds think alike – Tim Jan 14 '12 at 14:54
Thanks for the compliment... but a more appropriate notion is: folklore. – Did Jan 14 '12 at 15:07
Well, I was not aware of that it was folklore obviously. :-) – Tim Jan 14 '12 at 15:10
And the variant $g:(x,y)\mapsto xy/\sqrt{x^2+y^2}$ is useful when one looks for simple counterexamples about two-variables differentiability (instead of two-variables continuity here). – Did Jan 14 '12 at 15:18
Only consider joint differentiablility? That it is not defined and therefore not differentiable at (0,0) is what you meant? – Tim Jan 14 '12 at 15:24

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