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?


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 $t$.

  • $\begingroup$ +1 Great minds think alike $\endgroup$
    – Tim
    Jan 14 '12 at 14:54
  • 3
    $\begingroup$ Thanks for the compliment... but a more appropriate notion is: folklore. $\endgroup$
    – Did
    Jan 14 '12 at 15:07
  • $\begingroup$ Well, I was not aware of that it was folklore obviously. :-) $\endgroup$
    – Tim
    Jan 14 '12 at 15:10
  • 1
    $\begingroup$ 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). $\endgroup$
    – Did
    Jan 14 '12 at 15:18
  • $\begingroup$ Only consider joint differentiablility? That it is not defined and therefore not differentiable at (0,0) is what you meant? $\endgroup$
    – Tim
    Jan 14 '12 at 15:24

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