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

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!

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.
• 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. – user190080 Aug 8 '15 at 11:01
• 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 $|a_{mn}-L|<\epsilon$. – Tony Aug 8 '15 at 11:05
• 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 ($\pm$) – user190080 Aug 8 '15 at 11:20
• 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 $\mathop {\lim }\limits_{m \to \infty ,n \to \infty } {a_{mn}} = L$. – Tony Aug 8 '15 at 11:27
• 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 – user190080 Aug 8 '15 at 12:31
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 $$R$$.