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suppose, $f(x,y)$ is a bounded continuous function on $\mathbb{R}^2$. Consider $$\lim_{ y \rightarrow y' }\> \sup_{x \in \mathbb{R}} f(x,y).$$

In how far can you switch suprema and limits, or which possible term comes closest to switching these two? Ideally, there would be something like $$\sup_{x \in \mathbb{R}}\> \lim_{ y \rightarrow y' } f(x,y).$$

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You cannot in general. To take a very simple example, consider $f(x,y) = \sin(xy)$, and $y'=0$. Then we have: $$\lim_{y\to 0}\>\sup_{x\in\mathbb{R}}\sin(xy) = 1.$$ (Remember that if we take the limit as $y\to 0$, then we do not consider $y=0$, so $y\neq 0$ in $\sin(xy)$). But $$\sup_{x\in\mathbb{R}}\>\lim_{y\to 0}\sin(xy) = \sup_{x\in\mathbb{R}}\ 0 = 0.$$

However, if the limit of $f(x,y)$ as $y\to y'$ exists for every $x$, and the limit of the suprema exist, then you get one inequality: $$\sup_{x\in\mathbb{R}}\>\lim_{y\to y'}f(x,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}} f(x,y).$$ To see this, note that for each fixed $x_0\in\mathbb{R}$, $f(x_0,y) \leq \sup\limits_{x\in\mathbb{R}}f(x,y)$, so taking limits we have $$\lim_{y\to y'} f(x_0,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}}f(x,y).$$ Since this holds for each $x_0$, the supremum also satisfies the inequality, so $$\sup_{x\in\mathbb{R}}\>\lim_{y\to y'}f(x,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}}f(x,y).$$

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The first identity is false unless you use the definition: $\lim_{y\to y_0} g(y)=A$ if, and only if, $\forall\varepsilon>0\,\exists\delta>0(\, 0\lt|y-y_0|\lt\delta\Rightarrow|g(y)-A|<\varepsilon$). That is, if we skip $0\lt|y-y_0|$ the limit does not exist. –  AD. Dec 1 '10 at 19:07
    
@AD.: I'm not sure I follow; yes, I am using the fact that if we are taking a limit as $y\to 0$, then the value of $y$ is not equal to $0$. So each $\sup\sin(x,y)$ is being taken at a $y$ that is nonzero (which is why I get that the supremum is always $1$). Is that what you meant? –  Arturo Magidin Dec 1 '10 at 19:28
    
Yes, that is what I meant. I just thought it should be mentioned somewhere since it is not the standard definition :) –  AD. Dec 2 '10 at 4:56
    
@AD.: It's not? Granted, formally one also requires $y\in\mathrm{dom}(g)$, but what else? What is the standard definition you know? ("The nice thing about standards is that there are so many to choose from...") –  Arturo Magidin Dec 2 '10 at 5:01
    
Well, I think a more standard definition of limit goes: $\lim_{y\to y_0}g(y)=A$ iff $\forall\varepsilon>0\,\exists\delta>0(|y-y_0|<\delta\Rightarrow |g(y)-A|<\varepsilon)$ - in your case I would like to emphasise $\lim_{y\to y_0, y\ne y_0}$. –  AD. Dec 2 '10 at 5:20

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