Continuity of $f(x)$ on $[a,b]$ implies continuity of $f(t) = \displaystyle \min_{a \leq x \leq t}f(x)$ at [a,b].

Asked May 01 '19 at 13:01

Active May 01 '19 at 13:27

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Let $f(x)$ be a continuous function on $[a,b]$.

Show that $F(t) = \displaystyle \min_{a\leq x\leq t}f(x)$ is also continuous on $[a,b]$.

Statement seems obvious for me, especially when I try to draw some special cases of this problem, but I fail to derive a rigorous proof.

edited May 01 '19 at 13:27

asked May 01 '19 at 13:01

gadm

1

$\min_{a<x<t} f(x)$ is not defined at $t=a$. And in general the minimum need not exist anywhere at all. I suppose you mean $\le $ throughout – Hagen von Eitzen May 01 '19 at 13:06
Have you tried to write down what is it that you need to prove? A definition of continuity? Where do you get stuck? – uniquesolution May 01 '19 at 13:07
Also "$f:\Bbb R \to \Bbb R$" and "on $[a,b]$" are mutually exclusive. – May 01 '19 at 13:09
@Hagen, thanks, $\le$, of course. – gadm May 01 '19 at 13:20
@user658409 fixed. – gadm May 01 '19 at 13:23

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