$f:\left[ a,b\right] \rightarrow \mathbb{R}$ is a continuous function.$g\left( x\right) =\sup \left\{ f\left( t\right) :t\in \left[ a,x\right) \right\}$.How can I show that $g: \left[ a,b\right] \rightarrow \mathbb{R}$ is continuous?
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Don't you rather mean $t\in[a,x)$? – Hagen von Eitzen Aug 01 '17 at 11:19
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@HagenvonEitzen ı fixed – furkans Aug 01 '17 at 11:22
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1At least slightly related, and worth looking at even if it's not: the Rising Sun Lemma, https://en.wikipedia.org/wiki/Rising_sun_lemma. – John Hughes Aug 01 '17 at 11:23
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1$ g(a) = sup { f(t); t \in [a,a)}$? – Thiago Nascimento Aug 01 '17 at 12:33
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I guess you should start accepting some answers at this point and close the question. – Naive Aug 03 '17 at 18:01
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If $g(x)>f(x)$ then $f(y)<g(x)$ and hence $g(y)=g(x)$ for all $y$ sufficiently close to $x$.
If $g(x)=f(x)$, then for $y<x$, $f(y)\le g(y)\le g(x)$; and for $y>x$, there exists $z$ with $x<z\le y$ and $f(z)\ge g(y)$.
Hagen von Eitzen
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