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The supremum of a function $f(x)$, where $x$ is a member of the closed interval $[a,b]$, is one of the values assumed by $f(x)$ in the $x$-interval $[a,b]$

So, if $f(x)$ is piecewise continuous over an open interval of $x$, is sup $f(x)$ not one of the $f(x)$ values for $x \in (a,b)?$ Could anybody please explain why and show an example?

The intuition that I have (I may be very very wrong) is that, for an open interval $(a,b)$ of $x$, as $x\rightarrow a$ or $x\rightarrow b$, there are infinite $x$ values and $f(x)$ may be (asymptotically?) increasing, and $f(a)$ or $f(b)$ may be $> f(x_0)$ for all $a <x_0<b$.

Looking for some expert opinion.

Oussama Boussif
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Tuhin Paul
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2 Answers2

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The example $f(x) = x $ on the interval $0 < x < 1$ shows that the supremum may not be $f(a)$ for any $a$ in the interval.

But your question, literally, asks something different, and I'm going to answer that:

The function $f(x) = 1-x^2$ on the open interval $-1 < x < 1$ has a supremum of 1. (I should say $\sup \{ f(x) \mid -1 < x < 1 \} = 1$, I suppose).

But in this case the statement, "The supremum is not one of the $f(x)$ values for $a < x < b$," is wrong, because the supremum is $f(0)$, and $0$ is in the interval.

John Hughes
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Consider $f(x)= 1/x$ on $(0,1)$. You will not be able to find any $c$ in $(0,1)$ such that $f(c)= \sup \{f(x) :x \in (0,1) \}$

uniquesolution
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