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Let $\{ f_\alpha (x) \}_{\alpha \in I }$ a family of continuous functions from $[a,b]$ to $\Bbb{R}$ ($ a<b $). Now we know because $[a,b]$ is compact then every $f_\alpha$ is uniformly continuous and every $f_\alpha $ has maximum and minimum.

My questions are :

  1. Let $f(x) = \sup \{ f_\alpha (x) \} _{\alpha \in I }$ for $x \in [a,b] $ , is $f$ uniformly continuous ?

  2. Let $f(x) = \inf\{ f_\alpha (x) \} _{\alpha \in I }$ for $x \in [a,b] $ , is $f$ uniformly continuous ?

In the special case where we have $ k(x) = \max \{ f(x), g(x) \}$ and $u(x) = \min \{ f(x), g(x) \}$, then we know $k(x)$ and $u(x)$ are continuous functions and because $[a,b]$ is compact then $k(x)$ and $u(x)$ are uniformly continuous.

Watson
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amir bahadory
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1 Answers1

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This is wrong in general.

Define $$f_n : [0,1] \to [0,1] \qquad f_n(x) = \begin{cases} nx & 0≤x<\frac{1}{n}\\ 1 & \frac{1}{n}≤x<1-\frac{1}{n}\\ n-nx & 1-\frac{1}{n} ≤ x ≤ 1 \end{cases} $$

Here is a picture for $n=5$ :

function f_5

Then $f_n$ is uniformly continuous since it is continuous on a compact set. But $$f(x) = \sup\{f_n(x) \mid n ≥ 1\} = \begin{cases} 1 & 0<x<1\\ 0 & x=0,x=1\\ \end{cases}$$ is not even continuous.

You can find similar examples for the $\inf$. (Actually the result has no meaning in general, see for instance $f_n(x)=n$ since $\sup f_n = +\infty$ is not a real-valued function).

Watson
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