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For the continuous function $f : [a,b] \to \Bbb R$

Let $g(x) = \max\{f(y) : a \le y \le x \}. x\in[a,b]$ Show $g(x)$ is continuous

I tried to use $\epsilon\text{-}\delta$ first, but I failed. I thought that this function is continuous because of pasting lemma. Can you give me a few hints to show this.. ?

Michael Hardy
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fivestar
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  • Does this answer your question? https://math.stackexchange.com/questions/213682/is-the-maximum-function-of-a-continuous-function-continuous – Kavi Rama Murthy Jan 13 '20 at 08:29

1 Answers1

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Hint: Fix $x \in [a,b]$, let $I$ denote the interval $(x-\delta,x+\delta)$. Note that $$ \sup \{|g(y_1) - g(y_2)|: y_1,y_2 \in I\} \leq \sup \{|f(y_1) - f(y_2)|: y_1,y_2 \in I\} $$ and that by the triangle inequality, $$ \sup \{|f(y_1) - f(y_2)|: y_1,y_2 \in I\} \leq 2 \sup\{|f(y) - f(x)| : y \in I\}. $$ This should help you set up an $\epsilon$-$\delta$ proof.

Ben Grossmann
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