3
$\begingroup$

Let $f$ be a real-valued bounded function in $\mathbb{R}^{2}$ such that for all real $t,$ the functions $g_t(y)=f(t,y)$ and $h_t(x)=f(x,t)$ are non-decreasing. Then which of the following is/are correct?

  1. $f(x,x)$ is non-decreasing.
  2. $f$ can have uncountable number of discontinuities.
  3. $\lim\limits_{(x,y)\to(+\infty,+\infty)} f(x,y)$ exists.
  4. $\lim\limits_{(x,y)\to(+\infty,-\infty)} f(x,y)$ exists.

Option $1$ seems correct to me since $f$ is non-decreasing in each co-ordinate.

For 2) let $f(x,y)=\left\{\begin{align} 0, & \mbox{ if }~ x\leq0 \mbox{ or } y\leq0\\ 1, ~& x>0\mbox{ and y>0}\end{align}\right.$

Then $f$ is bounded in $\mathbb{R}^2,$ is non-decreasing and is discontinuous at each point on x-axis as well as on y-axis. So option 2) is also correct.

I am thinking that 3) and 4) should also be correct since $f$ is given to be bounded. However, I am not sure.

$\endgroup$
6
  • $\begingroup$ You should show more thinking. We can answer, but most users don't want to answer a blank post. E.g. you must have some ideas about 3), 4), and if you've been asked about 2) you've probably handled a similar case in class... so share all of the above $\endgroup$
    – FShrike
    Jan 3, 2023 at 17:52
  • $\begingroup$ @FShrike I have shared what I was thinking. $\endgroup$
    – Nik
    Jan 3, 2023 at 19:13
  • 1
    $\begingroup$ Your example for 2. is incorrect, as $g_{-1}(1)=f(-1,1)=1>0=f(-1,0)=g_{-1}(0).$ A correct example is f$(x,y)=0$ if $(x\le 0\lor y\le 0)$, and $f(x,y)=1$ if $(x>0\land y>0).$ $\endgroup$ Jan 4, 2023 at 1:03
  • 1
    $\begingroup$ For 1., if $x<y$ then $f(x,x)=g_x(x)\le g_x(y)=f(x,y)=h_y(x)\le h_y(y)=f(y,y).$ $\endgroup$ Jan 4, 2023 at 1:07
  • 1
    $\begingroup$ Concerning $4.$ you can take a function of the form $f(x,y)=g(x+y)$ where $g$ is strictly increasing, for example $g(t)=\arctan t.$ Then $f(x,-x)=g(0)$ and $f(x+1,-x)=g(1).$ $\endgroup$ Jan 5, 2023 at 21:59

1 Answer 1

1
$\begingroup$

The option $3.$ is correct. For $x\ge x_0$ and $y\ge y_0$ we have $$f(x,y)-f(x_0,y_0)=[f(x,y)-f(x,y_0)]+[f(x,y_0)-f(x_0,y_0)]\ge 0\ (*)$$ Denote $A=\displaystyle\sup_{x,y}f(x,y).$ We are going to show that $$\lim_{(x,y)\to (\infty,\infty)}f(x,y)=A$$ Let $\varepsilon >0.$ Then there exist $x_0,\ y_0,$ such that $$f(x_0,y_0)\ge A-\varepsilon $$ Then for $x\ge x_0,\ y\ge y_0,$ in view of $(*)$, we obtain $$A-\varepsilon \le f(x_0,y_0)\le f(x,y)\le A$$ Hence $$A-\varepsilon\le f(x,y)\le A,\quad x\ge x_0,\ y\ge y_0$$ This concludes the proof of the claim.

$\endgroup$
1
  • $\begingroup$ Nice proof. I thought it as : $f(x,x)$ is non-decreasing and bounded, so by monotone convergence theorem, $\lim\limits_{(x,y)\to\infty}f(x,y)$ must exist. $\endgroup$
    – Nik
    Jan 7, 2023 at 4:50

You must log in to answer this question.

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