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Suppose a sequence of continuous functions $f_n(x)$ converges to a function $f(x)$ on $x \in \mathcal{C}$ pointwise, i.e., $$\lim_{n \to \infty} f_{n}(x) = f(x), \ \forall \ x \in \mathcal{C}.$$ Then can we exchange the limit and max operations as follows

$$\lim_{n \to \infty} \max_{x \in \mathcal{C}} f_{n}(x) = \max_{x \in \mathcal{C}} f(x)$$ Or under what conditions of domain $\mathcal{C}$ does above hold? For example, how about $\mathcal{C} = \{x: x\geq 0\}$? How about $\mathcal{C}$ being a compact set?

Feng
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zxzx179
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1 Answers1

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False even when $C$ is compact. Let $C=[0,1]$. Let $f_n$ be $1$ and $[\frac 1 {2n}, \frac 1 n]$, $f(0)=0$ and $f(x)=0$ for $x >\frac 2 n$. Make $f$ piece-wise linear. Then $f_n$ is continuous for each $n$, $f_n(x) \to 0$ for each $x$ and $sup_x f_n(x)$ is $1$ for each $n$.

Of course the same example works for $[0,\infty)$ by just taking $f_n(x)=0$ for $x>1$.

Kavi Rama Murthy
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    It's a very nice example! Thanks so much! I guess we need to resort to some uniform convergence assumptions to guarantee the interchange to be true. – zxzx179 Aug 25 '19 at 00:36
  • Yes, when $C$ is finite this is true. @KarthikPN – Kavi Rama Murthy Jun 26 '20 at 10:33
  • But we do have $\lim_{n \to \infty} \max_{x \in \mathcal{C}} f_n(x) \ge \max_{x \in \mathcal{C}} f(x)$. See https://math.stackexchange.com/questions/1198366/exchanging-lim-and-inf?rq=1 – 1Rock Sep 04 '20 at 00:48