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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.2010-12-01
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    @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?2010-12-01
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    Yes, that is what I meant. I just thought it should be mentioned somewhere since it is not the standard definition :)2010-12-02
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    @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...")2010-12-02
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    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 $\lim_{y\to y_0, y\ne y_0}$.2010-12-02
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    I just took a look at wikipedia. Sure there are different standards --- 1-0 to you --- I rest my case.. :)2010-12-02
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    @AD. Really? Under that definition, a function with a removable discontinuity and is defined at $y_0$ has no limit at $y_0$ (because putting $y=y_0$ satisfies the antecedent but will not satisfy the consequent; I assume you consider the implication true for $y=y_0$ if $g$ is undefined at $y_0$). I've never seen the condition $y\neq y_0$ omitted. http://en.wikibooks.org/wiki/Calculus/Formal_Definition_of_the_Limit ; http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html; http://archives.math.utk.edu/visual.calculus/1/definition.6/index.html2010-12-02
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    @AD.: (cont) A few more. http://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit ; http://www.scottsarra.org/applets/calculus/EpsilonDelta.html ; http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/define_limit.html ; http://www.slu.edu/classes/maymk/Applets/EpsilonDelta.html ; http://www-math.mit.edu/~djk/18_01/chapter02/section02.html ; http://oregonstate.edu/instruct/mth251/cq/Stage3/Lesson/theDefinition.html ; http://www.math.uic.edu/~radford/math215f05/Hwk6Notes.pdf2010-12-02
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    @AD.: I did not mean to pile it on; I was cutting and pasting and did not see your comment until I left the page and came back... But it looks like at least 10-0 just from the web. I'll wager every one of my books in the office has the same stipulation.2010-12-02
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    Arturo, I am sorry - I was not thinking it through properly.. Of course you are right!2010-12-02