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Let $n\geq1$ and a sequence of upper semi-continuous functions $f_n:E \to \mathbb{R}$ where $E$ is some space (we can suppose it compact).

Let $f:E \to \mathbb{R}$ be an upper semi-continuous function and assume that $f_n \xrightarrow[n\to\infty]{} f$. Are there conditions I can add in order to say $$\sup_{E} f_n \xrightarrow[n\to\infty]{} \sup_{E} f \quad ?$$

Is an uniform convergence on $E$ helpful ?


marked as duplicate by Alex M., Yanior Weg, Shailesh, Cesareo, José Carlos Santos calculus May 19 at 16:55

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For every $x\in E$ you have $$f(x)=\lim_{n\to\infty}f_n(x)\le \limsup_{n\to\infty}\sup_E f_n.$$ Since this is true for every $x\in E$ you get $$\sup_E f \le \limsup_{n\to\infty}\sup_E f_n.$$ On the other hand, if $E$ is compact, since $f_n$ is upper semicontinuous there is $x_n\in E$ such that $f_n(x_n)=\sup_E f_n$.

Consider a subsequence $f_{n_k}$ such that $$\limsup_{n\to\infty}\sup_E f_n=\lim_{k\to\infty}\sup_E f_{n_k}.$$ By compactness of $E$, there is a further subsequence $x_{n_{k_i}}\to x_0 \in E$. By uniform convergence, $$\limsup_{n\to\infty}\sup_E f_n=\lim_{i\to\infty}\sup_E f_{n_{k_i}}=\lim_{i\to\infty}f_{n_{k_i}}(x_{n_{k_i}})=f(x_0)\le \sup_E f.$$ With a similar proof you can show that $$\sup_E f = \liminf_{n\to\infty}\sup_E f_n.$$ I used both compactness and uniform convergence.

  • $\begingroup$ Thanks a lot @Gio67 ! $\endgroup$ – anonymus Jul 18 '17 at 1:25

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