Is the sup of continuous functions still continuous?

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Suppose I have a family of functions $\{f_t, t\in [0, T]\}$, where $f_t:A\to\mathbb R$, with $A$ a generic set (not necessarily contained in $\mathbb R$). Suppose that, for all $t\in [0, T]$, $f_t$ is continuous in $x\in A$.

Is it true that $\sup_{t\in [0, T]}|f_t|$ is continuous in $x$?

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In general, you don't even know that $\sup_{t\in[0,T]} |f_t|$ even exists. For example, if $A=[0,1]$, you could have $f_t(x)=\frac{x}{t}$ for all $t>0$, and $f_0(x)=0$. In this case, the supremum doesn't exist.

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The answer is No. For example, let $f_t(x) = x^t$ for $0 \le x \le 1$ and $0 < t \le 1$ and let $f_0 = 0$. Then $\sup_t f_t(x) = \begin{cases} 1 \quad (0 < x \le 1) \\ 0 \quad (x = 0) \end{cases}$ and this function is not continuous.

However, $\sup_t f_t$ is still lower semicontinuous . More generally, suprema of families of lower semicontinuous functions are again lower semicontinuous under mild additional assumptions.