1

Let $f\colon [a,b] \to \mathbb R$ be continuous. Define $g\colon [a,b] \to \mathbb R$ by $g(x)=\max \{f(y); a\le y\le x\}$. Prove that $g$ is continuous on $(a,b)$.

question

I know $g(x)$ will continuous on $(a,b)$ but I do'nt know how to prove.. well. My ideais that While $f$ is non increasing, $g$ will be constant in that interval and when $f$ is increasing, $g$ is increasing too following $f$ and because constant function and $f$ are continuous, if we show that in critical point $g$ is also continuous, then the proof.will be.finished..?

I think that I am not used to proving with epsilon delta.. Anyway I want to know the proof with using epsilon delta or other

Martin Sleziak
  • 53,687
  • Note that you do not know that $g$ is differentiable, so you cannot talk about critical points. – parsiad Apr 15 '16 at 21:11
  • Right i meant that special point from which gx is not follow fx or follow and my idea was wrong because even if f is increasing, it can happen that gx is constant when previous maximum is greater –  Apr 15 '16 at 21:15
  • For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Apr 15 '16 at 23:20

1 Answers1

2

We want to show that $g$ is continuous at $x\in[a,b]$. Let $\epsilon>0$. By the continuity of $f$, there exists some $\delta>0$ such that for all $x^{\prime}\in[a,b]$, $|x-x^{\prime}|<\delta$ implies $|f(x) -f(x^{\prime})|<\epsilon$.

Let $x^\prime\in[a,b]$ such that $|x-x^{\prime}|<\delta$ and assume (without loss of generality) $x>x^\prime$. Note that $$ |g(x)-g(x^{\prime})|=|\max_{y\in[a,x]}f(y)-\max_{y\in[a,x^{\prime}]}f(y^{\prime})| $$ and there are two possibilities:

  1. The maximum of $f$ on $[a,x]$ occurs in $[a,x^{\prime}]$, in which case the above is zero.
  2. The maximum of $f$ on $[a,x]$ does not occur in $[a,x^{\prime}]$.

Can you finish the proof? Hint: the intermediate value theorem might be helpful.

Solution: In the second case, let $c$ denote a point in $(x^\prime,x]$ at which the maximum is attained. Since $f(c)>\max_{y\in[a,x^\prime]} f(y)$ and $f$ is continuous, we can use the intermediate value theorem to guarantee the existence of $d\in(x^{\prime},x]$ such that $\max_{y\in[a,x^{\prime}]}f(y)=f(d)$. Since $|c-d|<\delta$, it follows that $$|\max_{y\in[a,x]}f(y)-\max_{y\in[a,x^{\prime}]}f(y^{\prime})|=|f(c)-f(d)|<\epsilon.$$

parsiad
  • 25,154