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?
Continuity and Joint Continuity
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real-analysis
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0the difference is in definitions, so you may want to find an example what the function is continuous in each argument but not jointly – 2012-01-13
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0Also see an essentially same question [Does measurability/continuity of a mapping follow that of its sections?](http://math.stackexchange.com/q/96253/1281) – 2012-01-14
1 Answers
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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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0+1 [Great minds](http://math.stackexchange.com/a/96258/1281) [think alike](http://math.stackexchange.com/a/96259/1281) – 2012-01-14
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3Thanks for the compliment... but a more appropriate notion is: *folklore*. – 2012-01-14
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0Well, I was not aware of that it was folklore obviously. :-) – 2012-01-14
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1And 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). – 2012-01-14
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0Only consider joint differentiablility? That it is not defined and therefore not differentiable at (0,0) is what you meant? – 2012-01-14
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0So ''continuous'' = ''jointly continuous'' for several variables and otherwise it is called ''continuous in each argument''? – 2013-12-19