Suppose $f(x)$ is continuous on the closed interval $[a,b]$. Define $m(x)=\max_{a\leq s\leq x}\, f(s)$, $a\leq x\leq b$. Is $m(x)$ continuous necessarily? Thank you.
Is the maximum function of a continuous function continuous?
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0Yes it is. What are your thoughts? – 2012-10-14
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0Thank you. I know that $m(x)$ is increasing. So it is continuous unless it has jumps around some points. And this is not very possible since $f(x)$ is continuous. But I just do not know how to show this rigorously. Do you have any suggestion? – 2012-10-14
2 Answers
Yes it is.
HINT: Clearly $m(s)$ is monotonically non-decreasing on $[a,b]$. Thus, if it is discontinuous at some $x\in[a,b]$, it must have a jump discontinuity at $x$: either $\lim\limits_{s\to x^-}m(s) Suppose first that $\lim\limits_{s\to x^-}m(s) Then suppose that $m(x)<\lim\limits_{s\to x^+}m(s)$ for some $x\in[a,b)$. Show that $f(x)<\lim\limits_{s\to x^+}f(s)$, again contradicting the continuity of $f$. Added: Here’s a little more help with (1). Let $u=\lim\limits_{s\to x^-}m(s)$. For all $s\in[a,x)$ we have $$f(s)\le m(s)\le u (2) can be dealt with very similarly.
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0Thank you. That is a quite rigorous proof. It helps a lot. – 2012-10-14
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0@Knightgu: You’re welcome; glad it helps. – 2012-10-14
$f$ is uniformly continuous on $[a,b]$. Fix $\varepsilon>0$, choose $\delta$ such that $|f(x)-f(y)|\leq\varepsilon$ when $|x-y|\leq \delta$ and let $x\in [a,b]$. We have to show that $|m(x)-m(x+t)|\leq\varepsilon$ if $|t|\leq \delta$. We have for $t>0$ that
$$m(x+t)=\max_{0\leq s\leq x+t}f(s)=\max\{f(x),\max_{x\leq s\leq x+t}f(s)\},$$
and for $t<0$ that
$$m(x+t)=\max\{f(x-t),\max_{x-t\leq s\leq x}f(s)\}.$$
We have $|f(x)-\max_{x\leq s\leq x+t}f(s)\}|\leq\varepsilon$ when $0
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0I wonder if it is the case in your answer that $$m(x+t)=max\{m(x),\,max_{x\leq s\leq x+t}\,f(s)\}$$for t>0. Noting that $f(x)\leq m(x),$ we therefore have $$|m(x+t)-m(x)|\leq max\{0,\,|max_{x\leq t\leq x+t}\,f(s)-f(x)|\}.$$ The case $t<0$ is similar. – 2012-10-14