$\sup_{-\infty Does it mean the least upper bound of the set of $f(t)$ OR the least upper bound of $t$ which will then be applied to $f(t)$?
What supremum of a function
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real-analysis
1 Answers
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It means the least upper bound of the set $\{f(t) | t \in (-\infty,x) \}$.
To see the difference, consider $f(x) = \arctan (-x)$. $f$ is strictly monotone decreasing. Then you can see that $\sup \{t | t \in (-\infty,x) \} = x$, but $\sup \{f(t) | t \in (-\infty,x) \} = 1$.
This also illustrates that $\sup \{f(t) | t \in (-\infty,x) \} \neq f(\sup \{t | t \in (-\infty,x) \})$.
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0If it is of the set why it equals to $f(x-)$? – 2012-09-23
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0It is the least upper bound of the set, not the set. Also, it need not be $f(x-)$. – 2012-09-23
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0It is on p2 of "A Course in Probability Theory" is the word "monotonicity" mean anything here? – 2012-09-23
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0Yes, monotonicity is very relevant here. Monotone increasing means that if $t\leq t'$, then $f(t) \leq f(t')$. Similarly for decreasing, and if the $\leq$ is replaced by $<$, we have strict monotone increasing/decreasing. In this case, if $f$ is strictly monotone increasing, you will have $\sup \{f(t) | t \in (-\infty,x) \} = \lim_{t \to x} f(t)$. – 2012-09-23
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0So when will it equal to $f(x-)$? – 2012-09-23
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0Well, if $f$ is increasing then it will equal $f(x-) = \lim_{t \to x} f(t)$. – 2012-09-23
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0So when it is $f(x-)$ it is also f(sup t)? – 2012-09-23
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0If $f$ is continuous from the left. – 2012-09-23
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0p.s. In the book it assumed f is increasing $x_1 < x_2 \implies f(x_1) \leq f(x_2)$. So you are right. Thanks. – 2012-09-23