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Suppose $f(x,y)$ is defined on $[0,1]\times[0,1]$ and continuous on each dimension, i.e. $f(x,y_0)$ is continuous with respect to $x$ when fixing $y=y_0\in [0,1]$ and $f(x_0,y)$ is continuous with respect to $y$ when fixing $x=x_0\in [0,1]$. Show

$$\lim_{m \to \infty ,n \to \infty } f\left(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}\right) = f(x,y)$$

My attempt:

First, I know $$\lim\limits_{m \to \infty ,n \to \infty } \left(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}\right) = (x,y)$$

Secondly it looks $$\lim\limits_{m \to \infty }\lim\limits_{n \to \infty } f\left(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}\right) = \lim \limits_{m \to \infty } f\left(\frac{{\left\lfloor {mx} \right\rfloor }}{m},y\right) = f(x,y)$$

and $$\lim\limits_{n \to \infty } \lim\limits_{m \to \infty } f\left(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}\right) = \lim\limits_{n \to \infty } f\left(x,\frac{{\left\lfloor {ny} \right\rfloor }}{n}\right) = f(x,y)$$

since $f(x,y)$ is continuous on each dimension.

However, I am not sure if this can infer $\lim\limits_{m \to \infty ,n \to \infty } f(\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}) = f(x,y)$.

Can anyone provide some help? Thank you!

Added:

I am now sure $\lim\limits_{m \to \infty } \lim\limits_{n \to \infty } {a_{mn}} = \lim\limits_{n \to \infty } \lim\limits_{m \to \infty } {a_{mn}} = L$ does not imply $\lim\limits_{m \to \infty ,n \to \infty } {a_{mn}} =L$ in general. Hope someone can help solve the problem.

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  • $\begingroup$ how did you define $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } $, maybe something like $\lim_{(m,n) \to (\infty,\infty) }$? How did you prove $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } (\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}) = (x,y)$ and why do you think you cannot use this directly to conclude your claim? By the way, if all three limits exists, then they are equal. $\endgroup$ – user190080 Aug 8 '15 at 11:01
  • $\begingroup$ I don't think if the three limits exist then they are equal. Check this example. $f(x,y) = \begin{cases}0, & x = y = 0\\[2ex] \dfrac{xy}{x^2+y^2}, & x^2+y^2 > 0\end{cases}$. $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } a_{mn} = L$ means for any $\epsilon>0$ there exists $N\in \Bbb{N}$ st. $m,n>N$ implies $|a_{mn}-L|<\epsilon$. $\endgroup$ – Tony Aug 8 '15 at 11:05
  • $\begingroup$ I guess you mean the limiting point $(0,0)$? The problem here is that $\lim_{(x,y) \to (0,0) }$ does not exist - you can see this by approaching from the positive and negative side which leads to different results ($\pm$) $\endgroup$ – user190080 Aug 8 '15 at 11:20
  • $\begingroup$ Yes. Sorry about my imprecise comment. If $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } {a_{mn}}$ exists it can imply $\mathop {\lim }\limits_{m \to \infty } {a_{mn}}$ and the other when some conditions are met. But I don't think the converse is true. In my question, unfortunately, I only have $\mathop {\lim }\limits_{m \to \infty } \mathop {\lim }\limits_{n \to \infty } {a_{mn}} = \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } {a_{mn}} = L$, so I am unable to infer $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } {a_{mn}} = L$. $\endgroup$ – Tony Aug 8 '15 at 11:27
  • $\begingroup$ I am not quite sure I get your problem. You have proved that $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } (\frac{{\left\lfloor {mx} \right\rfloor }}{m},\frac{{\left\lfloor {ny} \right\rfloor }}{n}) = (x,y)$ holds, have you not? But this is already sufficient to conclude your claim (due to continuity)...please correct me if you think I took you completely wrong $\endgroup$ – user190080 Aug 8 '15 at 12:31
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It can't be done.For brevity, let $u=x-1/2,v=y-1/2$ and let $f(x,y)=uv/(u^2+v^2)$ when $u^2+v^2\not=0$, and $f(1/2,1/2)=0$. Then $f$ is continuous in each variable, but the formula whose limit you seek has the value $(m-1)(n-1)/4(m^2+n^2)$ when $x=y=1/2$ and $m,n$ are odd positive integers.This has no limit, e.g. try $m=n$ and then try $m=3n$. The trouble is that $f$ is not continuous as a function from $I^2$ to $R$.

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