# $\{f_n\}$ continuous such that $\inf f_n$ and $\sup f_n$ are not continuous

Find a sequence of continuous functions on $[0,1]$, $\{f_n\}$, such that $f(x):=\sup\{f_n(x):n\geq 1\}$ and $g(x):=\inf\{f_n(x):n\geq 1\}$ are both not continuous.

I kept finding examples where one was discontinuous but the other was continuous like $f_n(x)=|x|^\frac{1}{n}$ and $f_n(x)=x^n$

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You may consider $f_n(x) = \sin(nx)$. Then you can check that for all $n \ge 2$, $f(\frac{\pi}{2n}) = 1$ and $g(\frac{3\pi}{2n}) = -1$ (because $f_{n}(\frac{\pi}{2n}) = 1$, $f_n(\frac{3\pi}{2n}) = -1$, and clearly $f_k(x)$ is always between $-1$ and $1$). We also have $f(0) = g(0) = 0$. So neither $f$ nor $g$ is continuous at $0$.

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$f_n(x) = 1/2$ for $1/n < x < 1-1/n$, $f_n(0)=0, f_n(1)=1$, $f_n$ continuous and affine linear on $(0,1/n]$ and $[1-1/n,1)$. The $\inf$ is $1$ for $x=1$ but $=1/2$ for $x<1$ close to $1$. The $\sup$ is $0$ for $x=0$ but $=1/2$ for $x$ close to $0$, but not equal to zero.

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Let $(f_n)_{n\geq 1}$ be a sequence of continuous functions such that $f(x) = \sup\{f_n(x) \mid n \geq 1\}$ is not continuous, but is bounded, that is $|f(x)| < M$. Then $$g_n = \begin{cases} M + g_{\frac{n}{2}} & \text{ for } 2 | n \\ -M - g_{\frac{n+1}{2}} & \text{ otherwise} \end{cases}$$ is a sequence you are looking for. Clearly, for odd $n$ we have $g_n(x) < 0$ and $g_n(x) > 0$ for $n$ even. For this reason, for any $x$, the upper bound is $f(x)+M$ and the lower bound is $-f(x)-M$, and that completes the proof.

Hope that helps ;-)

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$$f_n:x\mapsto(1+\mathrm e^{n(x-a)})^{-1},\qquad a\in(0,1)$$

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