Assuming $f:[a,b]\to\mathbb{R}$ is continuous, how do you prove that the following function is continuous?
$$h(x)=\sup\{f(t): t\leq x\}$$
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What are your thoughts? - i.e. what have you tried so far? – Old John Jan 06 '14 at 22:59
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I have no idea, maybe some hint? – aiki93 Jan 06 '14 at 23:06
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Can you prove that the function $h$ is increasing? Do you know anything about possible discontinuities of monotone functions? – Andrés E. Caicedo Jan 06 '14 at 23:08
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See also: Prove functions defined by sup and inf are continuous and If $f$ is continuous on $[a,b]$ and $F(x) = \sup f([a,x])$. Prove that $F$ is continuous on $[a,b]$ . – Martin Sleziak Feb 27 '19 at 14:18
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@MartinSleziak I asked Jan 6 '14 at 22:57 – aiki93 Feb 27 '19 at 14:55
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@aiki93 One thing is that age is not the most important thing when closing as duplicates, older question can be closed as a duplicate of a newer one. There are many discussions about this on meta, for example Topics declared as duplicates in which order? and other questions linked there. – Martin Sleziak Feb 27 '19 at 15:42
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But the main reason I voted to close as a duplicate was that when I looked at the question it already was in the close votes review queue with two close votes with the close reason missing context - so it was very likely going to be closed anyway. To close as a duplicate seemed to me as a slightly better option. It is less harsh to the asker. And, moreover, the duplicate links directs users to other question(s) where the same thing is answered. – Martin Sleziak Feb 27 '19 at 16:07
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If $x<y$ and $|f(t)-f(x)|\le \delta$ for all $t\in [x,y]$ then $$ h(x) \le h(y) \le h(x)+\delta $$ since $$ h(y) = \max\{h(x),\sup\{f(t)\colon t \in [x,y]\}\} \le \max\{h(x),f(x)+\delta\} \le h(x)+\delta. $$
Emanuele Paolini
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