# Which of the following statements is/are correct about a real-valued bounded function in $\mathbb{R}^2?$

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.

• 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 Jan 3, 2023 at 17:52
• @FShrike I have shared what I was thinking.
– Nik
Jan 3, 2023 at 19:13
• 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 $(x>0\land y>0).$ Jan 4, 2023 at 1:03
• For 1., if $x<y$ then $f(x,x)=g_x(x)\le g_x(y)=f(x,y)=h_y(x)\le h_y(y)=f(y,y).$ Jan 4, 2023 at 1:07
• 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 $f(x+1,-x)=g(1).$ Jan 5, 2023 at 21:59

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.
• 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.