17
$\begingroup$

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?

$\endgroup$
2

1 Answer 1

21
$\begingroup$

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 $t$.

$\endgroup$
3
  • 2
    $\begingroup$ 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
    Commented Jan 14, 2012 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
    Commented Jan 14, 2012 at 15:24
  • 1
    $\begingroup$ So ''continuous'' = ''jointly continuous'' for several variables and otherwise it is called ''continuous in each argument''? $\endgroup$
    – blue
    Commented Dec 19, 2013 at 14:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .