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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).

Best

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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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    Arturo, I am sorry - I was not thinking it through prop$e$rly.. Of course you are right!2010-12-02