# Continuity in each of parameters mean joint continuity in the domain?

• Feb 15th 2013, 11:14 PM
bkarpuz
Continuity in each of parameters mean joint continuity in the domain?
Dear MHF members, here is my problem.
Suppose that we are given a real- or complex-valued function $\displaystyle f$
on a rectangular domain $\displaystyle D:=[a,b]\times[c,d]\subset\mathbb{R}^{2}$ such that
$\displaystyle f(t,\cdot)$ is continuous on $\displaystyle [c,d]$ for each fixed $\displaystyle t\in[a,b]$, and
$\displaystyle f(\cdot,s)$ is continuous on $\displaystyle [a,b]$ for each fixed $\displaystyle s\in[c,d]$.
• Does this mean that $\displaystyle f$ is continuous on $\displaystyle D$?
• If not, give a counter example, and provide additional
conditions on $\displaystyle f$ so that it continuous on $\displaystyle D$?

Thank you very much.
bkarpuz
• Feb 16th 2013, 01:23 AM
chiro
Re: Continuity in each of parameters mean joint continuity in the domain?
Hey bkarpuz.

You will have to use the delta epsilon mechanism to show its continuous but basically the limits should exist for all points in the region.

If you can show that the limit point of the function at certain values is a defined quantity then you're done. You should be able to directly use the continuity attributes to say lim x->a f(x,y0) = blah and lim y->b f(x0,y) = blah.
• Feb 16th 2013, 06:38 AM
bkarpuz
Re: Continuity in each of parameters mean joint continuity in the domain?
Thank you very much chiro, but I was looking for a more detailed explanation such as the following.
Separate continuity of $\displaystyle f$ does not mean joint continuity. But the converse is trivially true.
Consider the simple example
$\displaystyle f(t,s)=\begin{cases}\dfrac{ts}{t^{2}+s^{2}},&(t,s) \neq(0,0),\\ 0,&(t,s)=(0,0).\end{cases}$
Obviously, $\displaystyle f$ is separately continuous.
However, it is not joint continuous as the limit $\displaystyle (t,s)\to(0,0)$ does not exist.

The statement that separate continuity implies joint continuity turns out to be true
if separate continuity of $\displaystyle f$ is uniformly for at least one of the variables
(i.e., $\displaystyle \delta$ does not depent on the fixed parameter).
For details further see [1, Theorem 5, pp. 102].

Please correct me if I gave something wrong here.
Many thanks.
bkarpuz

References
[1] L. M. Graves, Theory of Functions of Real Variables, New York, 1946.
[2] V. Naik, Separately continuous functions (vs joint continuity) - YouTube