# Does $\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x)$?

Can you interchange limits and supremums of functions?

That is, does $$\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x) ?$$

Thank you!

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No, in general you cannot do that. Imagine if $X$ is $\mathbb{R}$ and $f_n(x)$ is zero except on $[n,n+1]$ where it is $1$. Work out both sides of the equation in this case...

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Thanks, awesome counterexample :D –  badatmath Nov 20 '11 at 5:53

No. Consider, for example $$f_n(x) = \frac{1}{(x-n)^2+1}$$ and $X=\mathbb R$. The supremum of each $f_n$ (and thus the limit of the suprema) is $1$, but the pointwise limit at each $x$ (and thus the supremum of the limits) is $0$.

You'll have better luck if you can assume that the $f_n$s converge uniformly on $X$.

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Thanks! Yeah I guess uniform convergence is a lot more powerful. –  badatmath Nov 20 '11 at 5:51

$$f_n(x)\leq \sup_{x\in A} f_n(x)$$ $$\Rightarrow {\liminf}_{n\rightarrow \infty} f_n(x)\leq {\liminf}_{n\rightarrow \infty}\sup_{x\in A}f_n(x)$$ $$\Rightarrow \sup_{x\in A}{\liminf}_{n\rightarrow \infty} f_n(x)\leq {\liminf}_{n\rightarrow \infty}\sup_{x\in A}f_n(x)$$

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